sparsh35/aops

1id,link,no,problem,solution,answer21,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_1,1,"What is the ones digit of \[222,222-22,222-2,222-222-22-2?\] $\textbf{(A) } 0\qquad\textbf{(B) } 2\qquad\textbf{(C) } 4\qquad\textbf{(D) } 6\qquad\textbf{(E) } 8$","We can rewrite the expression as \[222,222-(22,222+2,222+222+22+2).\]3We note that the units digit of the addition is $0$ because all the units digits of the five numbers are $2$ and $5*2=10$ , which has a units digit of $0$4Now, we have something with a units digit of $0$ subtracted from $222,222$ . The units digit of this expression is obviously $2$ , and we get $\boxed{2}$ as our answer.",252,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_1,2,"What is the ones digit of \[222,222-22,222-2,222-222-22-2?\] $\textbf{(A) } 0\qquad\textbf{(B) } 2\qquad\textbf{(C) } 4\qquad\textbf{(D) } 6\qquad\textbf{(E) } 8$","222,222-22,222 = 200,0006200,000 - 2,222 = 1977787197778 - 222 = 1975568197556 - 22 = 1975349197534 - 2 = 195753210So our answer is $\boxed{2}$",2113,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_1,3,"What is the ones digit of \[222,222-22,222-2,222-222-22-2?\] $\textbf{(A) } 0\qquad\textbf{(B) } 2\qquad\textbf{(C) } 4\qquad\textbf{(D) } 6\qquad\textbf{(E) } 8$","We only care about the unit's digits.12Thus, $2-2$ ends in $0$ $0-2$ ends in $8$ $8-2$ ends in $6$ $6-2$ ends in $4$ , and $4-2$ ends in $\boxed{2}$",2134,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_1,4,"What is the ones digit of \[222,222-22,222-2,222-222-22-2?\] $\textbf{(A) } 0\qquad\textbf{(B) } 2\qquad\textbf{(C) } 4\qquad\textbf{(D) } 6\qquad\textbf{(E) } 8$","We just take the units digit of each and subtract, or you can do it this way by adding an extra ten to the first number (so we don't get a negative number): \[12-2-(2+2+2+2)=10-8=2\] Thus, we get the answer $\boxed{2}$",2145,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_3,1,"Four squares of side length $4, 7, 9,$ and $10$ are arranged in increasing size order so that their left edges and bottom edges align. The squares alternate in color white-gray-white-gray, respectively, as shown in the figure. What is the area of the visible gray region in square units? [asy]  size(150); filldraw((0,0)--(10,0)--(10,10)--(0,10)--cycle,gray(0.7),linewidth(1)); filldraw((0,0)--(9,0)--(9,9)--(0,9)--cycle,white,linewidth(1)); filldraw((0,0)--(7,0)--(7,7)--(0,7)--cycle,gray(0.7),linewidth(1)); filldraw((0,0)--(4,0)--(4,4)--(0,4)--cycle,white,linewidth(1)); draw((11,0)--(11,4),linewidth(1)); draw((11,6)--(11,10),linewidth(1)); label(""$10$"",(11,5),fontsize(14pt)); draw((10.75,0)--(11.25,0),linewidth(1)); draw((10.75,10)--(11.25,10),linewidth(1)); draw((0,11)--(3,11),linewidth(1)); draw((5,11)--(9,11),linewidth(1)); draw((0,11.25)--(0,10.75),linewidth(1)); draw((9,11.25)--(9,10.75),linewidth(1)); label(""$9$"",(4,11),fontsize(14pt)); draw((-1,0)--(-1,1),linewidth(1)); draw((-1,3)--(-1,7),linewidth(1)); draw((-1.25,0)--(-0.75,0),linewidth(1)); draw((-1.25,7)--(-0.75,7),linewidth(1)); label(""$7$"",(-1,2),fontsize(14pt)); draw((0,-1)--(1,-1),linewidth(1)); draw((3,-1)--(4,-1),linewidth(1)); draw((0,-1.25)--(0,-.75),linewidth(1)); draw((4,-1.25)--(4,-.75),linewidth(1)); label(""$4$"",(2,-1),fontsize(14pt)); [/asy] $\textbf{(A)}\ 42 \qquad \textbf{(B)}\ 45\qquad \textbf{(C)}\ 49\qquad \textbf{(D)}\ 50\qquad \textbf{(E)}\ 52$",We work inwards. The area of the outer shaded square is the area of the whole square minus the area of the second largest square. The area of the inner shaded region is the area of the third largest square minus the area of the smallest square. The sum of these areas is \[10^2 - 9^2 + 7^2 - 4^2 = 19 + 33 = \boxed{52}\],52156,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_3,2,"Four squares of side length $4, 7, 9,$ and $10$ are arranged in increasing size order so that their left edges and bottom edges align. The squares alternate in color white-gray-white-gray, respectively, as shown in the figure. What is the area of the visible gray region in square units? [asy]  size(150); filldraw((0,0)--(10,0)--(10,10)--(0,10)--cycle,gray(0.7),linewidth(1)); filldraw((0,0)--(9,0)--(9,9)--(0,9)--cycle,white,linewidth(1)); filldraw((0,0)--(7,0)--(7,7)--(0,7)--cycle,gray(0.7),linewidth(1)); filldraw((0,0)--(4,0)--(4,4)--(0,4)--cycle,white,linewidth(1)); draw((11,0)--(11,4),linewidth(1)); draw((11,6)--(11,10),linewidth(1)); label(""$10$"",(11,5),fontsize(14pt)); draw((10.75,0)--(11.25,0),linewidth(1)); draw((10.75,10)--(11.25,10),linewidth(1)); draw((0,11)--(3,11),linewidth(1)); draw((5,11)--(9,11),linewidth(1)); draw((0,11.25)--(0,10.75),linewidth(1)); draw((9,11.25)--(9,10.75),linewidth(1)); label(""$9$"",(4,11),fontsize(14pt)); draw((-1,0)--(-1,1),linewidth(1)); draw((-1,3)--(-1,7),linewidth(1)); draw((-1.25,0)--(-0.75,0),linewidth(1)); draw((-1.25,7)--(-0.75,7),linewidth(1)); label(""$7$"",(-1,2),fontsize(14pt)); draw((0,-1)--(1,-1),linewidth(1)); draw((3,-1)--(4,-1),linewidth(1)); draw((0,-1.25)--(0,-.75),linewidth(1)); draw((4,-1.25)--(4,-.75),linewidth(1)); label(""$4$"",(2,-1),fontsize(14pt)); [/asy] $\textbf{(A)}\ 42 \qquad \textbf{(B)}\ 45\qquad \textbf{(C)}\ 49\qquad \textbf{(D)}\ 50\qquad \textbf{(E)}\ 52$",We can calculate it as the sum of the areas of $2$ smaller trapezoids and $2$ larger trapezoids. \[2\left(\cfrac{(7+4)(7-4)}{2}\right)+2\left(\cfrac{(10+9)(10-9)}{2}\right)=10^2 - 9^2 + 7^2 - 4^2 = 19 + 33 = \boxed{52}\],52167,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_4,1,"When Yunji added all the integers from $1$ to $9$ , she mistakenly left out a number. Her incorrect sum turned out to be a square number. What number did Yunji leave out?17$\textbf{(A) } 5\qquad\textbf{(B) } 6\qquad\textbf{(C) } 7\qquad\textbf{(D) } 8\qquad\textbf{(E) } 9$","The sum of the digits from $1-9$ are $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$ . Note that one of the answer choices (that is equal to Yunju's digits) subtracted from her sum of $45$ must equal a square.18Note that $6^2 = 36$ is a very close square to the sum of 45. Checking, we see that $45 - 9 = 36 = 6^2$ works.19Therefore, the missing number is $\boxed{9}$",9208,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_5,1,"Aaliyah rolls two standard 6-sided dice. She notices that the product of the two numbers rolled is a multiple of $6$ . Which of the following integers cannot be the sum of the two numbers?21$\textbf{(A) } 5\qquad\textbf{(B) } 6\qquad\textbf{(C) } 7\qquad\textbf{(D) } 8\qquad\textbf{(E) } 9$","First, figure out all pairs of numbers whose product is 6. Then, using the process of elimination, we can find the following:22$\textbf{(A)}$ is possible: $2\times 3$23$\textbf{(C)}$ is possible: $1\times 6$24$\textbf{(D)}$ is possible: $2\times 6$25$\textbf{(E)}$ is possible: $3\times 6$26The only integer that cannot be the sum is $\boxed{6}.$",6279,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_7,1,"$3 \times 7$ rectangle is covered without overlap by 3 shapes of tiles: $2 \times 2$ $1\times4$ , and $1\times1$ , shown below. What is the minimum possible number of $1\times1$ tiles used?28$\textbf{(A) } 1\qquad\textbf{(B)} 2\qquad\textbf{(C) } 3\qquad\textbf{(D) } 4\qquad\textbf{(E) } 5$","We can eliminate B, C, and D, because they are not $21$ subtracted by any multiple of $4$ . Finally, we see that there is no way to have A, so the solution is $\boxed{5}$",52910,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_7,2,"$3 \times 7$ rectangle is covered without overlap by 3 shapes of tiles: $2 \times 2$ $1\times4$ , and $1\times1$ , shown below. What is the minimum possible number of $1\times1$ tiles used?30$\textbf{(A) } 1\qquad\textbf{(B)} 2\qquad\textbf{(C) } 3\qquad\textbf{(D) } 4\qquad\textbf{(E) } 5$","Let $x$ be the number of $1x1$ tiles. There are $21$ squares and each $2x2$ or $1x4$ tile takes up 4 squares, so $x \equiv 1 \pmod{4}$ , so it is either $1$ or $5$ . Color the columns, starting with red, then blue, and alternating colors, ending with a red column. There are $12$ red squares and $9$ blue squares, but each $2x2$ and $1x4$ shape takes up an equal number of blue and red squares, so there must be $3$ more $1x1$ tiles on red squares than on blue squares, which is impossible if there is just one, so the answer is $\boxed{5}$ , which can easily be confirmed to work.",53111,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_7,3,"$3 \times 7$ rectangle is covered without overlap by 3 shapes of tiles: $2 \times 2$ $1\times4$ , and $1\times1$ , shown below. What is the minimum possible number of $1\times1$ tiles used?32$\textbf{(A) } 1\qquad\textbf{(B)} 2\qquad\textbf{(C) } 3\qquad\textbf{(D) } 4\qquad\textbf{(E) } 5$","Suppose there are $a$ different $2\times 2$ tiles, $b$ different $4\times 1$ tiles and $c$ different $1\times 1$ tiles. Since the areas of these tiles must total up to $21$ (area of the whole grid), we have \[4a + 4b + c = 21.\] Reducing modulo $4$ gives $c\equiv 1\pmod{4}$ , or $c = 1$ or $c = 5$33If $c = 1$ , then $a + b = 5$ . After some testing, there is no valid pair $(a, b)$ that works, so the answer must be $\boxed{5}$ , which can be constructed in many ways.",53412,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_8,1,"On Monday Taye has $$2$ . Every day, he either gains $$3$ or doubles the amount of money he had on the previous day. How many different dollar amounts could Taye have on Thursday, $3$ days later?35$\textbf{(A) } 3\qquad\textbf{(B) } 4\qquad\textbf{(C) } 5\qquad\textbf{(D) } 6\qquad\textbf{(E) } 7$","How many values could be on the first day? Only $2$ dollars. The second day, you can either add $3$ dollars, or double, so you can have $5$ dollars, or $4$ . For each of these values, you have $2$ values for each. For $5$ dollars, you have $10$ dollars or $8$ , and for $4$ dollars, you have $8$ dollars or $ $7$ . Now, you have $2$ values for each of these. For $10$ dollars, you have $13$ dollars or $20$ , for $8$ dollars, you have $16$ dollars or $11$ , for $8$ dollars, you have $16$ dollars or $11$ , and for $7$ dollars, you have $14$ dollars or $10$36On the final day, there are 11, 11, 16, and 16 repeating, leaving you with $8-2 = \boxed{6}$ different values.",63713,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_8,2,"On Monday Taye has $$2$ . Every day, he either gains $$3$ or doubles the amount of money he had on the previous day. How many different dollar amounts could Taye have on Thursday, $3$ days later?38$\textbf{(A) } 3\qquad\textbf{(B) } 4\qquad\textbf{(C) } 5\qquad\textbf{(D) } 6\qquad\textbf{(E) } 7$","Continue as in Solution 1 to get $7$ $8$ , or $10$ dollars by the 2nd day. The only way to get the same dollar amount occurring twice by branching (multiply by $2$ or adding $3$ ) from here is if $7+3=10\cdot 2$ or $7+3=8\cdot 2$ which both aren't true. Hence our answer is $3\cdot2=\boxed{6}$",63914,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_9,1,"All the marbles in Maria's collection are red, green, or blue. Maria has half as many red marbles as green marbles and twice as many blue marbles as green marbles. Which of the following could be the total number of marbles in Maria's collection?40$\textbf{(A) } 24\qquad\textbf{(B) } 25\qquad\textbf{(C) } 26\qquad\textbf{(D) } 27\qquad\textbf{(E) } 28$","Since she has half as many red marbles as green, we can call the number of red marbles $x$ , and the number of green marbles $2x$ .41Since she has half as many green marbles as blue, we can call the number of blue marbles $4x$ . 42Adding them up, we have $7x$ marbles. The number of marbles therefore must be a multiple of $7$ . The only possible answer is $\boxed{28}.$",284315,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_9,2,"All the marbles in Maria's collection are red, green, or blue. Maria has half as many red marbles as green marbles and twice as many blue marbles as green marbles. Which of the following could be the total number of marbles in Maria's collection?44$\textbf{(A) } 24\qquad\textbf{(B) } 25\qquad\textbf{(C) } 26\qquad\textbf{(D) } 27\qquad\textbf{(E) } 28$","Suppose Maria has $g$ green marbles and let $t$ be the total number of marbles. She then has $\frac{g}{2}$ red marbles and $2g$ blue marbles. Altogether, Maria has \[g + \frac{g}{2} + 2g = \frac{7g}{2} = t\] marbles, implying that $g = \dfrac{2t}{7},$ so $t$ must be a multiple of $7$ . The only multiple of $7$ is $\boxed{28}.$",284516,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_10,1,"In January $1980$ the Mauna Loa Observatory recorded carbon dioxide $(CO2)$ levels of $338$ ppm (parts per million). Over the years the average $CO2$ reading has increased by about $1.515$ ppm each year. What is the expected $CO2$ level in ppm in January $2030$ ? Round your answer to the nearest integer.46$\textbf{(A)}\ 399\qquad \textbf{(B)}\ 414\qquad \textbf{(C)}\ 420\qquad \textbf{(D)}\ 444\qquad \textbf{(E)}\ 459$","This is a time period of $50$ years, so we can expect the ppm to increase by $50*1.515=75.75~76$ $76+338=\boxed{414}$",4144717,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_10,2,"In January $1980$ the Mauna Loa Observatory recorded carbon dioxide $(CO2)$ levels of $338$ ppm (parts per million). Over the years the average $CO2$ reading has increased by about $1.515$ ppm each year. What is the expected $CO2$ level in ppm in January $2030$ ? Round your answer to the nearest integer.48$\textbf{(A)}\ 399\qquad \textbf{(B)}\ 414\qquad \textbf{(C)}\ 420\qquad \textbf{(D)}\ 444\qquad \textbf{(E)}\ 459$","For each year that has passed, the ppm will increase by $1.515$ . In $2030$ , the CO2 would have increased by $50\cdot 1.515 \approx. 76,$ so the total ppm of CO2 will be $76 + 338 = \boxed{414}.$",4144918,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_10,3,"In January $1980$ the Mauna Loa Observatory recorded carbon dioxide $(CO2)$ levels of $338$ ppm (parts per million). Over the years the average $CO2$ reading has increased by about $1.515$ ppm each year. What is the expected $CO2$ level in ppm in January $2030$ ? Round your answer to the nearest integer.50$\textbf{(A)}\ 399\qquad \textbf{(B)}\ 414\qquad \textbf{(C)}\ 420\qquad \textbf{(D)}\ 444\qquad \textbf{(E)}\ 459$","2030 - 1980 = 50 years.51338 + 50 * 1.515 = 338 + 75.75 = 413.75 for 2030 ppm level $=\boxed{414}$",4145219,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_11,1,"The coordinates of $\triangle ABC$ are $A(5,7)$ $B(11,7)$ , and $C(3,y)$ , with $y>7$ . The area of $\triangle ABC$ is 12. What is the value of $y$53[asy]  draw((3,11)--(11,7)--(5,7)--(3,11));  dot((5,7)); label(""$A(5,7)$"",(5,7),S);  dot((11,7)); label(""$B(11,7)$"",(11,7),S);  dot((3,11)); label(""$C(3,y)$"",(3,11),NW);  [/asy]54$\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad \textbf{(E) }12$","The triangle has base $6,$ which means its height satisfies \[\dfrac{6h}{2}=3h=12.\] This means that $h=4,$ so the answer is $7+4=\boxed{11}$",115520,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_11,2,"The coordinates of $\triangle ABC$ are $A(5,7)$ $B(11,7)$ , and $C(3,y)$ , with $y>7$ . The area of $\triangle ABC$ is 12. What is the value of $y$56[asy]  draw((3,11)--(11,7)--(5,7)--(3,11));  dot((5,7)); label(""$A(5,7)$"",(5,7),S);  dot((11,7)); label(""$B(11,7)$"",(11,7),S);  dot((3,11)); label(""$C(3,y)$"",(3,11),NW);  [/asy]57$\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad \textbf{(E) }12$","[asy] size(10cm); draw((5,7)--(11,7)--(3,11)--cycle); draw((3,11)--(3,7)--(5,7),red); draw((3,7.5)--(3.5,7.5)--(3.5,7)); label(""$A(5,7)$"", (5,7),S); label(""$B(11,7)$"", (11,7),S); label(""$C(3,y)$"", (3,11),W); label(""$D(3,7)$"", (3,7),SW); [/asy] Label point $D(3,7)$ as the point at which $CD\perp DA$ . We now have $[\triangle ABC] = [\triangle BCD] - [\triangle ACD]$ , where the brackets denote areas. On the right hand side, both of these triangles are right, so we can just compute the two sides of each triangle. The two side lengths of $\triangle ACD$ are $y-7$ and $5-3=2$ . The two side lengths of $\triangle BCD$ are $y-7$ and $11-3 = 8.$ Now,58\[[\triangle ABC] = 12  = \frac{1}{2}\cdot (y-7)\cdot 8 - \frac{1}{2}\cdot (y-7)\cdot 2  = 3(y-7)\]59Dividing by $3$ gives $y -7 = 4,$ so $y = \boxed{11}.$",116021,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_11,3,"The coordinates of $\triangle ABC$ are $A(5,7)$ $B(11,7)$ , and $C(3,y)$ , with $y>7$ . The area of $\triangle ABC$ is 12. What is the value of $y$61[asy]  draw((3,11)--(11,7)--(5,7)--(3,11));  dot((5,7)); label(""$A(5,7)$"",(5,7),S);  dot((11,7)); label(""$B(11,7)$"",(11,7),S);  dot((3,11)); label(""$C(3,y)$"",(3,11),NW);  [/asy]62$\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad \textbf{(E) }12$","By the Shoelace Theorem, $\triangle ABC$ has area \[\frac{1}{2}|(y \cdot 11 + 7 \cdot 5 + 7 \cdot 3) - (3 \cdot 7 + 11 \cdot 7 + 5 \cdot y)| = \frac{1}{2}|(11y + 56) - (98 + 5y)| = \frac{1}{2}|6y - 42|\] . From the problem, this is equal to $12$ . We now solve for y.63$\frac{1}{2}|6y - 42| = 12$64$|6y-42| = 24$65$6y - 42 = 24$ OR $6y - 42 = -24$66$6y = 66$ OR $6y = 18$67$y = 11$ OR $y = 3$68However, since, as stated in the problem, $y > 7$ , our only valid solution is $\boxed{11}$",116922,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_11,4,"The coordinates of $\triangle ABC$ are $A(5,7)$ $B(11,7)$ , and $C(3,y)$ , with $y>7$ . The area of $\triangle ABC$ is 12. What is the value of $y$70[asy]  draw((3,11)--(11,7)--(5,7)--(3,11));  dot((5,7)); label(""$A(5,7)$"",(5,7),S);  dot((11,7)); label(""$B(11,7)$"",(11,7),S);  dot((3,11)); label(""$C(3,y)$"",(3,11),NW);  [/asy]71$\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad \textbf{(E) }12$","As in the figure, the triangle is determined by the vectors $\begin{bmatrix}-2 \\ y-7\end{bmatrix}$ and $\begin{bmatrix}6\\0\end{bmatrix}$ . Recall that the absolute value of the determinant of these vectors is the area of the parallelogram determined by those vectors; the triangle has half the area of that parallelogram. Then we must have that $\frac{1}{2}|\begin{vmatrix}-2 & y-7\\6 & 0\end{vmatrix}| = 12 \implies \begin{vmatrix}-2 & y-7\\6 & 0\end{vmatrix} = \pm 24$ . Expanding the determinants, we find that $-6(y-7) = 24$ or $-6(y-7) = -24$ . Solving each equation individually, we find that $y = 3$ or $y = 11$ . However, the problem states that $y > 7$ , so the only valid solution is $\boxed{11}$",117223,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_12,1,"Rohan keeps a total of 90 guppies in 4 fish tanks.73How many guppies are in the 4th tank?74$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 23 \qquad \textbf{(D)}\ 24 \qquad \textbf{(E)}\ 26$","Let $x$ denote the number of guppies in the first tank.75Then, we have the following for the number of guppies in the rest of the tanks:76The number of guppies in all of the tanks combined is 90, so we can write the equation77$x + x + 1 + x + 1 + 2 + x + 1 + 2 + 3 = 90$78Simplifying the equation gives79$4x + 10 = 90$80Solving the resulting equation gives $x = 20$ , so the number of guppies in the fourth tank is $20 + 1 + 2 + 3 = 26$81Therefore, the correct answer is $\boxed{26}$",268224,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_12,2,"Rohan keeps a total of 90 guppies in 4 fish tanks.83How many guppies are in the 4th tank?84$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 23 \qquad \textbf{(D)}\ 24 \qquad \textbf{(E)}\ 26$","Suppose there are no guppies in the first tank.85Then, the number of guppies in the other tanks are $1,3,$ and $6,$ or $10$ guppies in total.86We need to add $90 - 10 = 80$ guppies into $4$ tanks or $20$ guppies in each tank.87So the number of guppies in the fourth tank is $20 + 6 = \boxed{26}.$",268825,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_13,1,"Buzz Bunny is hopping up and down a set of stairs, one step at a time. In how many ways can Buzz start on the ground, make a sequence of $6$ hops, and end up back on the ground?89(For example, one sequence of hops is up-up-down-down-up-down.)902024-AMC8-q13.png91$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 12$","Looking at the answer choices, you see that you can list them out. 92Doing this gets you:93$UUDDUD$94$UDUDUD$95$UUUDDD$96$UDUUDD$97$UUDUDD$98Counting all the paths listed above gets you $\boxed{5}$",59926,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_13,2,"Buzz Bunny is hopping up and down a set of stairs, one step at a time. In how many ways can Buzz start on the ground, make a sequence of $6$ hops, and end up back on the ground?100(For example, one sequence of hops is up-up-down-down-up-down.)1012024-AMC8-q13.png102$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 12$","Any combination can be written as some re-arrangement of $UUUDDD$ . Clearly we must end going down, and start going up, so we need the number of ways to insert 2 $U$ 's and 2 $D$ 's into $U\, \_ \, \_ \, \_ \, \_ \, D$ . There are ${4\choose 2}=6$ ways, but we have to remove the case $UDDUUD$ , giving us $\boxed{5}$",510327,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_13,4,"Buzz Bunny is hopping up and down a set of stairs, one step at a time. In how many ways can Buzz start on the ground, make a sequence of $6$ hops, and end up back on the ground?104(For example, one sequence of hops is up-up-down-down-up-down.)1052024-AMC8-q13.png106$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 12$","First step is U, last step is D.107After third step we can get only positions 3 or 1.108In the first case there is only one way UUUDDD.109In the second case we have two way to get this position UDU and UUD.110Similarly, we have two way return to position 0 (UDD and DUD).111Therefore, we have $1 + 2 \cdot 2 = \boxed{5}$",511228,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_13,5,"Buzz Bunny is hopping up and down a set of stairs, one step at a time. In how many ways can Buzz start on the ground, make a sequence of $6$ hops, and end up back on the ground?113(For example, one sequence of hops is up-up-down-down-up-down.)1142024-AMC8-q13.png115$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 12$","We can find the total cases then deduct the ones that don't work.116Let $U$ represent ""Up"" and $D$ represent ""Down"". We know that in order to land back at the bottom of the stairs, we must have an equal number of $U$ 's and $D$ 's, therefore six hops means $3$ of each.117The number of ways to arrange $3$ $U$ 's and $3$ $D$ 's is $\dfrac{6!}{(3!)^2}=\dfrac{720}{36}=20$118Case $1$ : Start with $D$119Case $2$ : Start with $UDD$120Case $3$ : Start with $UUDDD$121Case $4$ : Start with $UDUDD$122Case $1$ is asking us how many ways there are to arrange $3$ $U$ 's and $2$ $D$ 's, which is $\dfrac{5!}{3!2!}=\dfrac{120}{12}=10$123Case $2$ is asking us how many ways there are to arrange $2$ $U$ 's and $1$ $D$ , which is $\dfrac{3!}{2!1!}=\dfrac{6}{2}=3$124Case $3$ is asking us how many ways there are to arrange $1$ $U$ , which is $1$125Case $4$ is asking us the same thing as Case $3$ , giving us $1$126Therefore, deducting all cases from $20$ gives $20-10-3-1-1=\boxed{5}$",512729,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_14,1,"The one-way routes connecting towns $A,M,C,X,Y,$ and $Z$ are shown in the figure below(not drawn to scale).The distances in kilometers along each route are marked. Traveling along these routes, what is the shortest distance from A to Z in kilometers?1282024-AMC8-q14.png129$\textbf{(A)}\ 28 \qquad \textbf{(B)}\ 29 \qquad \textbf{(C)}\ 30 \qquad \textbf{(D)}\ 31 \qquad \textbf{(E)}\ 32$","We can simply see that path $A \rightarrow X \rightarrow M \rightarrow Y \rightarrow C \rightarrow Z$ will give us the smallest value. Adding, $5+2+6+5+10 = \boxed{28}$ . This is nice as it’s also the smallest value, solidifying our answer.",2813030,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_14,2,"The one-way routes connecting towns $A,M,C,X,Y,$ and $Z$ are shown in the figure below(not drawn to scale).The distances in kilometers along each route are marked. Traveling along these routes, what is the shortest distance from A to Z in kilometers?1312024-AMC8-q14.png132$\textbf{(A)}\ 28 \qquad \textbf{(B)}\ 29 \qquad \textbf{(C)}\ 30 \qquad \textbf{(D)}\ 31 \qquad \textbf{(E)}\ 32$","We can execute Dijkstra's algorithm by hand to find the shortest path from $A$ to every other town, including $Z$ . This effectively proves that, assuming we execute the algorithm correctly, that we will have found the shortest distance. The distance estimates for each step of the algorithm (from $A$ to each node) are shown below:133\[\begin{array}{|c|c|c|c|c|c|c|} \hline \text{Current node} & A & M & C & X & Y & Z \\ \hline A & 0 & 8 & \infty & 5 & \infty & \infty \\ X & 0 & 7 & \infty & 5 & 15 & \infty \\ M & 0 & 7 & 21 & 5 & 13 & 32 \\ Y & 0 & 7 & 18 & 5 & 13 & 30 \\ C & 0 & 7 & 18 & 5 & 13 & 28 \\ Z & 0 & 7 & 18 & 5 & 13 & \textbf{28} \\ \hline \end{array}\] The steps are as follows: starting with the initial node $A$ , set $d(A)=0$ and $d(v)=\infty$ for all $v \in \{M,C,X,Y,Z\}$ where $d(v)$ indicates the distance from $A$ to $v$ . Consider the outgoing edges $(A,X)$ and $(A,M)$ and update the distance estimates $d(X)=5$ and $d(M)=8$ , completing the first row of the table.134The node $X$ is the unvisited node with the lowest distance estimate, so we will consider $X$ and its outgoing edges $(X,Y)$ and $(X,M)$ . The distance estimate $d(Y)$ equals $d(X)+10=15$ , and the distance estimate $d(M)$ updates to $d(X)+2=7$ , because $7 < 8$ . This completes the second row of the table. Repeating this process for each unvisited node (in order of its distance estimate) yields the correct distance $d(Z) = \boxed{28}$ once the algorithm is complete.",2813531,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_14,3,"The one-way routes connecting towns $A,M,C,X,Y,$ and $Z$ are shown in the figure below(not drawn to scale).The distances in kilometers along each route are marked. Traveling along these routes, what is the shortest distance from A to Z in kilometers?1362024-AMC8-q14.png137$\textbf{(A)}\ 28 \qquad \textbf{(B)}\ 29 \qquad \textbf{(C)}\ 30 \qquad \textbf{(D)}\ 31 \qquad \textbf{(E)}\ 32$","From $A$ , we want to find the shortest route to $Z$ , so we want to try to find the shortest path through each node (not necessarily all of them). We should follow the arrows, since all of them lead to $Z$ . From $A$ , there are $2$ paths we can take, to $M$ $(8)$ , or to $X$ $(5)$ . We travel to $X$ , since $5 < 8$ . From $X$ , we go to $M$ $(2 < 10)$ , we go to $Y$ $6 < 14 < 25$ , we go to $C$ $(5 < 17)$ and finally go to $Z$ . Adding up gives $5 + 2 + 6 + 5 + 10 = 28 = \boxed{28}$",2813832,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_14,4,"The one-way routes connecting towns $A,M,C,X,Y,$ and $Z$ are shown in the figure below(not drawn to scale).The distances in kilometers along each route are marked. Traveling along these routes, what is the shortest distance from A to Z in kilometers?1392024-AMC8-q14.png140$\textbf{(A)}\ 28 \qquad \textbf{(B)}\ 29 \qquad \textbf{(C)}\ 30 \qquad \textbf{(D)}\ 31 \qquad \textbf{(E)}\ 32$","We can cross off a few routes:141Finally, we are left with a single path AXMYCZ from A to Z which adding it up gives $28 = \boxed{28}.$",2814233,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_15,1,"Let the letters $F$ $L$ $Y$ $B$ $U$ $G$ represent distinct digits. Suppose $\underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y}$ is the greatest number that satisfies the equation143\[8\cdot\underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y}=\underline{B}~\underline{U}~\underline{G}~\underline{B}~\underline{U}~\underline{G}.\]144What is the value of $\underline{F}~\underline{L}~\underline{Y}+\underline{B}~\underline{U}~\underline{G}$145$\textbf{(A)}\ 1089 \qquad \textbf{(B)}\ 1098 \qquad \textbf{(C)}\ 1107 \qquad \textbf{(D)}\ 1116 \qquad \textbf{(E)}\ 1125$","The highest that $FLYFLY$ can be would have to be $124124$ , and it cannot be higher than that because then it would exceed the $6$ -digit limit set on $BUGBUG$146So, if we start at $124124\cdot8$ , we get $992992$ , which would be wrong because both $B \& U$ would be $9$ , and the numbers cannot be repeated between different letters.147If we move on to the next highest, $123123$ , and multiply by $8$ , we get $984984$ . All the digits are different, so $FLY+BUG$ would be $123+984$ , which is $1107$ . So, the answer is $\boxed{1107}$",10714834,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_15,2,"Let the letters $F$ $L$ $Y$ $B$ $U$ $G$ represent distinct digits. Suppose $\underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y}$ is the greatest number that satisfies the equation149\[8\cdot\underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y}=\underline{B}~\underline{U}~\underline{G}~\underline{B}~\underline{U}~\underline{G}.\]150What is the value of $\underline{F}~\underline{L}~\underline{Y}+\underline{B}~\underline{U}~\underline{G}$151$\textbf{(A)}\ 1089 \qquad \textbf{(B)}\ 1098 \qquad \textbf{(C)}\ 1107 \qquad \textbf{(D)}\ 1116 \qquad \textbf{(E)}\ 1125$","Notice that $\underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y} = 1000(\underline{F}~\underline{L}~\underline{Y}) + \underline{F}~\underline{L}~\underline{Y}$152Likewise, $\underline{B}~\underline{U}~\underline{G}~\underline{B}~\underline{U}~\underline{G} = 1000(\underline{B}~\underline{U}~\underline{G}) + \underline{B}~\underline{U}~\underline{G}$153Therefore, we have the following equation:154$8 \times 1001(\underline{F}~\underline{L}~\underline{Y}) = 1001(\underline{B}~\underline{U}~\underline{G})$155Simplifying the equation gives156$8(\underline{F}~\underline{L}~\underline{Y}) = (\underline{B}~\underline{U}~\underline{G})$157We can now use our equation to test each answer choice.158We have that $123123 \times 8 = 984984$ , so we can find the sum:159$\underline{F}~\underline{L}~\underline{Y}+\underline{B}~\underline{U}~\underline{G} = 123 + 984 = 1107$160So, the correct answer is $\boxed{1107}$",10716135,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_15,3,"Let the letters $F$ $L$ $Y$ $B$ $U$ $G$ represent distinct digits. Suppose $\underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y}$ is the greatest number that satisfies the equation162\[8\cdot\underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y}=\underline{B}~\underline{U}~\underline{G}~\underline{B}~\underline{U}~\underline{G}.\]163What is the value of $\underline{F}~\underline{L}~\underline{Y}+\underline{B}~\underline{U}~\underline{G}$164$\textbf{(A)}\ 1089 \qquad \textbf{(B)}\ 1098 \qquad \textbf{(C)}\ 1107 \qquad \textbf{(D)}\ 1116 \qquad \textbf{(E)}\ 1125$","Note that $FLY+BUG = 9 \cdot FLY$ . Thus, we can check the answer choices and find $FLY$ through each of the answer choices, we find the 1107 works, so the answer is $\boxed{1107}$",10716536,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_16,2,"Minh enters the numbers $1$ through $81$ into the cells of a $9 \times 9$ grid in some order. She calculates the product of the numbers in each row and column. What is the least number of rows and columns that could have a product divisible by $3$166$\textbf{(A) } 8\qquad\textbf{(B) } 9\qquad\textbf{(C) } 10\qquad\textbf{(D) } 11\qquad\textbf{(E) } 12$","Note you can swap/rotate any configuration of rows, such that all the rows and columns that have a product of 3 are in the top left. Hence the points are bounded by a $a \times b$ rectangle. This has $ab$ area and $a+b$ rows and columns divisible by $3$ . We want $ab\ge 27$ and $a+b$ minimized.167If $ab=27$ , we achieve minimum with $a+b=9+3=12$168If $ab=28$ ,our best is $a+b=7+4=11$ . Note if $a+b=10$ , then $ab\le 25$ , and hence there is no smaller answer, and we get $\boxed{11}$",1116937,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_16,3,"Minh enters the numbers $1$ through $81$ into the cells of a $9 \times 9$ grid in some order. She calculates the product of the numbers in each row and column. What is the least number of rows and columns that could have a product divisible by $3$170$\textbf{(A) } 8\qquad\textbf{(B) } 9\qquad\textbf{(C) } 10\qquad\textbf{(D) } 11\qquad\textbf{(E) } 12$","For a row or column to have a product divisible by $3$ , there must be a multiple of $3$ in the row or column. To create the least amount of rows and columns with multiples of $3$ , we must find a way to keep them all together, to minimize the total number of rows and columns. From $1$ to $81$ , there are $27$ multiples of $3$ $81/3$ ). So we have to fill $27$ cells with numbers that are multiples of $3$ . If we put $25$ of these numbers in a $5 x 5$ grid, there would be $5$ rows and $5$ columns ( $10$ in total), with products divisible by $3$ . However, we have $27$ numbers, so $2$ numbers remain to put in the $9 x 9$ grid. If we put both numbers in the $6$ th column, but one in the first row, and one in the second row, (next to the $5 x 5$ already filled), we would have a total of $6$ columns now, and still $5$ rows with products that are multiples of $3$ . So the answer is $\boxed{11}$",1117138,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_16,4,"Minh enters the numbers $1$ through $81$ into the cells of a $9 \times 9$ grid in some order. She calculates the product of the numbers in each row and column. What is the least number of rows and columns that could have a product divisible by $3$172$\textbf{(A) } 8\qquad\textbf{(B) } 9\qquad\textbf{(C) } 10\qquad\textbf{(D) } 11\qquad\textbf{(E) } 12$","In the numbers $1$ to $81$ , there are 27 multiples of three. In order to minimize the rows and columns, the best way is to make a square. However, the closest square is $25$ , meaning there are two multiples of three remaining. However, you can place these multiples right above the 5x5 square, meaning the answer is $\boxed{11}$ (Error making remote request. No response to HTTP request) ~ e___",1117339,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_18,1,"Three concentric circles centered at $O$ have radii of $1$ $2$ , and $3$ . Points $B$ and $C$ lie on the largest circle. The region between the two smaller circles is shaded, as is the portion of the region between the two larger circles bounded by central angle $BOC$ , as shown in the figure below. Suppose the shaded and unshaded regions are equal in area. What is the measure of $\angle{BOC}$ in degrees?174[asy] size(150); import graph;  draw(circle((0,0),3)); real radius = 3; real angleStart = -54;  // starting angle of the sector real angleEnd = 54;  // ending angle of the sector label(""$O$"",(0,0),W); pair O = (0, 0); filldraw(arc(O, radius, angleStart, angleEnd)--O--cycle, gray); filldraw(circle((0,0),2),gray); filldraw(circle((0,0),1),white); draw((1.763,2.427)--(0,0)--(1.763,-2.427)); label(""$B$"",(1.763,2.427),NE); label(""$C$"",(1.763,-2.427),SE);  [/asy] $\textbf{(A) } 108\qquad\textbf{(B) } 120\qquad\textbf{(C) } 135\qquad\textbf{(D) } 144\qquad\textbf{(E) } 150$","Let $x=\angle{BOC}$175We see that the shaded region is the inner ring plus a sector $x^\circ$ of the outer ring. The area of this in terms of $x$ is $\left( 4 \pi - \pi \right)+\frac{x}{360} \left( 9 \pi - 4 \pi \right)$ . This simplifies to $3 \pi + \frac{x}{360}(5 \pi)$176Also, the unshaded portion is comprised of the smallest circle plus the sector $(360-x)^\circ$ of the outer ring. The area of this is $\pi + \frac{360-x}{360}(5 \pi)$177We are told these are equal, therefore $\pi + \frac{x}{360}(5 \pi) = 3 \pi + \frac{360-x}{360}(5 \pi)$ . Solving for $x$ reveals $x=\boxed{108}$",10817840,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_18,2,"Three concentric circles centered at $O$ have radii of $1$ $2$ , and $3$ . Points $B$ and $C$ lie on the largest circle. The region between the two smaller circles is shaded, as is the portion of the region between the two larger circles bounded by central angle $BOC$ , as shown in the figure below. Suppose the shaded and unshaded regions are equal in area. What is the measure of $\angle{BOC}$ in degrees?179[asy] size(150); import graph;  draw(circle((0,0),3)); real radius = 3; real angleStart = -54;  // starting angle of the sector real angleEnd = 54;  // ending angle of the sector label(""$O$"",(0,0),W); pair O = (0, 0); filldraw(arc(O, radius, angleStart, angleEnd)--O--cycle, gray); filldraw(circle((0,0),2),gray); filldraw(circle((0,0),1),white); draw((1.763,2.427)--(0,0)--(1.763,-2.427)); label(""$B$"",(1.763,2.427),NE); label(""$C$"",(1.763,-2.427),SE);  [/asy] $\textbf{(A) } 108\qquad\textbf{(B) } 120\qquad\textbf{(C) } 135\qquad\textbf{(D) } 144\qquad\textbf{(E) } 150$","Notice for the 3rd most outer ring of the circle, the ratio of the shaded region to non-shaded region is the ratio of $\angle{BOC}$ to $360-\angle{BOC}$ . With that, all we need to do is solve for the shaded region.180The inner most circle has radius $1$ , and the second circle has radius 2. Therefore, the first shaded area has $4 \pi - \pi = 3 \pi$ area. The circle has total area $9 \pi$ , so the other shaded region must have $1.5 \pi$ area, as the non-shaded and shaded area is equivalent. So for the 3rd outer ring, the total area is $9 \pi - 4 \pi = 5 \pi$ , so the non-shaded part of the outer ring is $5 \pi - 1.5 \pi = 3.5 \pi$181Now as said before, the ratio of these two areas is the ratio of $\angle{BOC}$ and $360 - \angle{BOC}$ . So, $\frac{3.5}{1.5} = \frac{7}{3}$ . We have $7x:3x$ where $7x+3x = 360$ $x = 36$ , so our answer is $3x = 108, \boxed{108}$",10818241,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_18,3,"Three concentric circles centered at $O$ have radii of $1$ $2$ , and $3$ . Points $B$ and $C$ lie on the largest circle. The region between the two smaller circles is shaded, as is the portion of the region between the two larger circles bounded by central angle $BOC$ , as shown in the figure below. Suppose the shaded and unshaded regions are equal in area. What is the measure of $\angle{BOC}$ in degrees?183[asy] size(150); import graph;  draw(circle((0,0),3)); real radius = 3; real angleStart = -54;  // starting angle of the sector real angleEnd = 54;  // ending angle of the sector label(""$O$"",(0,0),W); pair O = (0, 0); filldraw(arc(O, radius, angleStart, angleEnd)--O--cycle, gray); filldraw(circle((0,0),2),gray); filldraw(circle((0,0),1),white); draw((1.763,2.427)--(0,0)--(1.763,-2.427)); label(""$B$"",(1.763,2.427),NE); label(""$C$"",(1.763,-2.427),SE);  [/asy] $\textbf{(A) } 108\qquad\textbf{(B) } 120\qquad\textbf{(C) } 135\qquad\textbf{(D) } 144\qquad\textbf{(E) } 150$","The AMC 8s allow a ruler tool that you can rotate and drag. You can use the tool to make a straight segment (which we know is $180$ degrees), and we let the angle of desire be $x$ . We can estimate that $180-x$ is just about $30$ degrees short of $x$ itself, so $x-30=180-x$ , solving gives $x=105$ , therefore the closest answer is $\boxed{108}$",10818442,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_18,4,"Three concentric circles centered at $O$ have radii of $1$ $2$ , and $3$ . Points $B$ and $C$ lie on the largest circle. The region between the two smaller circles is shaded, as is the portion of the region between the two larger circles bounded by central angle $BOC$ , as shown in the figure below. Suppose the shaded and unshaded regions are equal in area. What is the measure of $\angle{BOC}$ in degrees?185[asy] size(150); import graph;  draw(circle((0,0),3)); real radius = 3; real angleStart = -54;  // starting angle of the sector real angleEnd = 54;  // ending angle of the sector label(""$O$"",(0,0),W); pair O = (0, 0); filldraw(arc(O, radius, angleStart, angleEnd)--O--cycle, gray); filldraw(circle((0,0),2),gray); filldraw(circle((0,0),1),white); draw((1.763,2.427)--(0,0)--(1.763,-2.427)); label(""$B$"",(1.763,2.427),NE); label(""$C$"",(1.763,-2.427),SE);  [/asy] $\textbf{(A) } 108\qquad\textbf{(B) } 120\qquad\textbf{(C) } 135\qquad\textbf{(D) } 144\qquad\textbf{(E) } 150$","Suppose the desired angle is some fraction $x$ of the total degree measure of the circle. We now compile a list of the shaded and unshaded areas. The inner circle of radius $1$ is completely unshaded, so it contributes $1$ to the unshaded area. (Everything will be a multiple of $\pi$ , so we omit it.) The inner annulus has area $2^2 - 1^2 = 3$ , which it contributes to the shaded area. The outer annulus has a total area of $3^2 - 2^2 = 5$ ; the fraction $x$ is shaded, so the shaded portion of the outer annulus contributes $5x$ to the shaded area, while the other $1 - x$ fraction is unshaded, so the unshaded portion contributes $5(1-x)$ to the unshaded area. We now equate and solve. \[1 + 5(1-x) = 3 + 5x\] Upon solving, we find that $x = \frac{3}{10}$ , so the degree measure is $360 \cdot \frac{3}{10} = \boxed{108}$",10818643,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_20,2,"Any three vertices of the cube $PQRSTUVW$ , shown in the figure below, can be connected to form a triangle. (For example, vertices $P$ $Q$ , and $R$ can be connected to form isosceles $\triangle PQR$ .) How many of these triangles are equilateral and contain $P$ as a vertex?187[asy] unitsize(4); pair P,Q,R,S,T,U,V,W; P=(0,30); Q=(30,30); R=(40,40); S=(10,40); T=(10,10); U=(40,10); V=(30,0); W=(0,0); draw(W--V); draw(V--Q); draw(Q--P); draw(P--W); draw(T--U); draw(U--R); draw(R--S); draw(S--T); draw(W--T); draw(P--S); draw(V--U); draw(Q--R); dot(P); dot(Q); dot(R); dot(S); dot(T); dot(U); dot(V); dot(W); label(""$P$"",P,NW); label(""$Q$"",Q,NW); label(""$R$"",R,NE); label(""$S$"",S,N); label(""$T$"",T,NE); label(""$U$"",U,NE); label(""$V$"",V,SE); label(""$W$"",W,SW); [/asy]188$\textbf{(A)}0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3 \qquad \textbf{(E) }6$","Each other compatible point must be an even number of edges away from P, so the compatible points are R, V, and T. Therefore, we must choose two of the three points, because P must be a point in the triangle. So, the answer is ${3 \choose 2} = \boxed{3}$",318944,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_20,3,"Any three vertices of the cube $PQRSTUVW$ , shown in the figure below, can be connected to form a triangle. (For example, vertices $P$ $Q$ , and $R$ can be connected to form isosceles $\triangle PQR$ .) How many of these triangles are equilateral and contain $P$ as a vertex?190[asy] unitsize(4); pair P,Q,R,S,T,U,V,W; P=(0,30); Q=(30,30); R=(40,40); S=(10,40); T=(10,10); U=(40,10); V=(30,0); W=(0,0); draw(W--V); draw(V--Q); draw(Q--P); draw(P--W); draw(T--U); draw(U--R); draw(R--S); draw(S--T); draw(W--T); draw(P--S); draw(V--U); draw(Q--R); dot(P); dot(Q); dot(R); dot(S); dot(T); dot(U); dot(V); dot(W); label(""$P$"",P,NW); label(""$Q$"",Q,NW); label(""$R$"",R,NE); label(""$S$"",S,N); label(""$T$"",T,NE); label(""$U$"",U,NE); label(""$V$"",V,SE); label(""$W$"",W,SW); [/asy]191$\textbf{(A)}0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3 \qquad \textbf{(E) }6$","List them out- you get $PRV$ $PRT$ , and $PVT$ . Therefore, the answer is $\boxed{3}$",319245,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_21,1,"A group of frogs (called an army) is living in a tree. A frog turns green when in the shade and turns yellow193when in the sun. Initially, the ratio of green to yellow frogs was $3 : 1$ . Then $3$ green frogs moved to the194sunny side and $5$ yellow frogs moved to the shady side. Now the ratio is $4 : 1$ . What is the difference195between the number of green frogs and the number of yellow frogs now?196$\textbf{(A) } 10\qquad\textbf{(B) } 12\qquad\textbf{(C) } 16\qquad\textbf{(D) } 20\qquad\textbf{(E) } 24$","Since the original ratio is $3:1$ and the new ratio is $4:1$ , the number of frogs must be a multiple of $12$ , the only solutions left are $(B)$ and $(E)$197Let's start with $12$ frogs:198We must have $9$ frogs in the shade and $3$ frogs in the sun. After the change, there would be $11$ frogs in the shade and $1$ frog in the sun, which is not a $4:1$ ratio.199Therefore the answer is: $\boxed{24}$",2420046,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_22,1,"A roll of tape is $4$ inches in diameter and is wrapped around a ring that is $2$ inches in diameter. A cross section of the tape is shown in the figure below. The tape is $0.015$ inches thick. If the tape is completely unrolled, approximately how long would it be? Round your answer to the nearest $100$ inches.201$\textbf{(A) } 300\qquad\textbf{(B) } 600\qquad\textbf{(C) } 1200\qquad\textbf{(D) } 1500\qquad\textbf{(E) } 1800$","The roll of tape is $1/0.015=$ 66 layers thick. In order to find the total length, we have to find the average of each concentric circle and multiply it by $66$ . Since the diameter of the small circle is $2$ inches and the diameter of the large one is $4$ inches, the ""middle value"" is $3$ . Therefore, the average circumference is $3\pi$ . Multiplying $3\pi \cdot 66$ gives $(B) \boxed{600}$",60020247,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_22,2,"A roll of tape is $4$ inches in diameter and is wrapped around a ring that is $2$ inches in diameter. A cross section of the tape is shown in the figure below. The tape is $0.015$ inches thick. If the tape is completely unrolled, approximately how long would it be? Round your answer to the nearest $100$ inches.203$\textbf{(A) } 300\qquad\textbf{(B) } 600\qquad\textbf{(C) } 1200\qquad\textbf{(D) } 1500\qquad\textbf{(E) } 1800$","There are about $\dfrac{1}{0.015}=\dfrac{200}{3}$ ""full circles"" of tape, and with average circumference of $\dfrac{4+2}{2}\pi=3\pi.$ $\dfrac{200}{3} \cdot 3\pi=200\pi,$ which means the answer is $\boxed{600}$",60020448,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_22,3,"A roll of tape is $4$ inches in diameter and is wrapped around a ring that is $2$ inches in diameter. A cross section of the tape is shown in the figure below. The tape is $0.015$ inches thick. If the tape is completely unrolled, approximately how long would it be? Round your answer to the nearest $100$ inches.205$\textbf{(A) } 300\qquad\textbf{(B) } 600\qquad\textbf{(C) } 1200\qquad\textbf{(D) } 1500\qquad\textbf{(E) } 1800$","The volume of the tape is always the same, but we can either calculate it when the tape is unrolled as a really long, thin rectangular prism, or we can calculate it as a cylinder with a hole cut out of it. When we calculate it as a long rectangular prism, we can say that the length is $X$ (this is what the problem wants!) and the width is $Y$ . Then, the volume is, of course, $0.015 \cdot X \cdot Y.$ Now, notice that the ""width"" of our rectangular prism is also the ""height"" of our cylinder with a hole cut out of it. Then, we can calculate the volume as base times height, or in this case, $3\pi \cdot Y.$ Now, since the volume always stays the same, we know that $3\pi \cdot Y = 0.015 \cdot X \cdot Y.$ Cancelling the $Y$ 's give us an equation for $X$ , and if we approximate $\pi$ as $3$ , then $X = \boxed{600}$ . Yay!",60020649,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_22,4,"A roll of tape is $4$ inches in diameter and is wrapped around a ring that is $2$ inches in diameter. A cross section of the tape is shown in the figure below. The tape is $0.015$ inches thick. If the tape is completely unrolled, approximately how long would it be? Round your answer to the nearest $100$ inches.207$\textbf{(A) } 300\qquad\textbf{(B) } 600\qquad\textbf{(C) } 1200\qquad\textbf{(D) } 1500\qquad\textbf{(E) } 1800$","If you cannot notice that the average diameter is $3$ , you can still solve this problem by the following method.208The same with solution 1, we have $\frac{1000}{0.015}$ layers of tape. If we consider every layers with the diameter $2$ , the length should be $\frac{1000}{0.015}2\pi\approx 400$ . If the diameter is seem as $4$ , the length should be $800$ . So, the length is between $400$ and $800$ , the only possible answer is $\boxed{600}$",60020950,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_23,2,"Rodrigo has a very large sheet of graph paper. First he draws a line segment connecting point $(0,4)$ to point $(2,0)$ and colors the $4$ cells whose interiors intersect the segment, as shown below. Next Rodrigo draws a line segment connecting point $(2000,3000)$ to point $(5000,8000)$ . How many cells will he color this time?210[asy]  filldraw((0,4)--(1,4)--(1,3)--(0,3)--cycle, gray(.75), gray(.5)+linewidth(1)); filldraw((0,3)--(1,3)--(1,2)--(0,2)--cycle, gray(.75), gray(.5)+linewidth(1)); filldraw((1,2)--(2,2)--(2,1)--(1,1)--cycle, gray(.75), gray(.5)+linewidth(1)); filldraw((1,1)--(2,1)--(2,0)--(1,0)--cycle, gray(.75), gray(.5)+linewidth(1));  draw((-1,5)--(-1,-1),gray(.9)); draw((0,5)--(0,-1),gray(.9)); draw((1,5)--(1,-1),gray(.9)); draw((2,5)--(2,-1),gray(.9)); draw((3,5)--(3,-1),gray(.9)); draw((4,5)--(4,-1),gray(.9)); draw((5,5)--(5,-1),gray(.9));  draw((-1,5)--(5, 5),gray(.9)); draw((-1,4)--(5,4),gray(.9)); draw((-1,3)--(5,3),gray(.9)); draw((-1,2)--(5,2),gray(.9)); draw((-1,1)--(5,1),gray(.9)); draw((-1,0)--(5,0),gray(.9)); draw((-1,-1)--(5,-1),gray(.9));   dot((0,4)); label(""$(0,4)$"",(0,4),NW);  dot((2,0)); label(""$(2,0)$"",(2,0),SE);  draw((0,4)--(2,0));  draw((-1,0) -- (5,0), arrow=Arrow); draw((0,-1) -- (0,5), arrow=Arrow);  [/asy]211$\textbf{(A) } 6000\qquad\textbf{(B) } 6500\qquad\textbf{(C) } 7000\qquad\textbf{(D) } 7500\qquad\textbf{(E) } 8000$","Draw a line in the lattice which from $(2,3)$ to $(5,8)$ , notice that the line crossed 7 blocks in this pattern. Such a pattern is repeated 1000 times between $(2000,3000)$ and $(5000,8000)$ , then the answer is $\boxed{7000}$",021251,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_24,1,"Jean has made a piece of stained glass art in the shape of two mountains, as shown in the figure below. One mountain peak is $8$ feet high while the other peak is $12$ feet high. Each peak forms a $90^\circ$ angle, and the straight sides form a $45^\circ$ angle with the ground. The artwork has an area of $183$ square feet. The sides of the mountain meet at an intersection point near the center of the artwork, $h$ feet above the ground. What is the value of $h?$213[asy] unitsize(.3cm); filldraw((0,0)--(8,8)--(11,5)--(18,12)--(30,0)--cycle,gray(0.7),linewidth(1)); draw((-1,0)--(-1,8),linewidth(.75)); draw((-1.4,0)--(-.6,0),linewidth(.75)); draw((-1.4,8)--(-.6,8),linewidth(.75)); label(""$8$"",(-1,4),W); label(""$12$"",(31,6),E); draw((-1,8)--(8,8),dashed); draw((31,0)--(31,12),linewidth(.75)); draw((30.6,0)--(31.4,0),linewidth(.75)); draw((30.6,12)--(31.4,12),linewidth(.75)); draw((31,12)--(18,12),dashed); label(""$45^{\circ}$"",(.75,0),NE,fontsize(10pt)); label(""$45^{\circ}$"",(29.25,0),NW,fontsize(10pt)); draw((8,8)--(7.5,7.5)--(8,7)--(8.5,7.5)--cycle); draw((18,12)--(17.5,11.5)--(18,11)--(18.5,11.5)--cycle); draw((11,5)--(11,0),dashed); label(""$h$"",(11,2.5),E); [/asy]214$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 4\sqrt{2} \qquad \textbf{(D)}\ 5\sqrt{2} \qquad \textbf{(E)}\ 6$","Extend the ""inner part"" of the mountain so that the image is two right triangles that overlap in a third right triangle as shown. [asy] unitsize(.2cm); draw((0,0)--(8,8)--(11,5)--(18,12)--(30,0)--cycle,linewidth(1)); draw((11,5)--(6, 0)--(16, 0)--cycle,linewidth(0.5)); label(""$8\sqrt{2}$"",(4,4),NW); label(""$12\sqrt{2}$"",(24,6),NE); draw((8,8)--(7.5,7.5)--(8,7)--(8.5,7.5)--cycle); draw((18,12)--(17.5,11.5)--(18,11)--(18.5,11.5)--cycle); draw((11,5)--(11,0),dashed); label(""$h$"",(11,2.5),E); [/asy] The side length of the largest right triangle is $12\sqrt{2},$ which means its area is $144.$ Similarly, the area of the second largest right triangle is $64$ (the side length is $8\sqrt{2}$ ), and the area of the overlap is $h^2$ (the side length is $h\sqrt{2}$ ). Thus, \[144+64-h^2=183,\] which means that the answer is $\boxed{5}.$",521552,https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_24,2,"Jean has made a piece of stained glass art in the shape of two mountains, as shown in the figure below. One mountain peak is $8$ feet high while the other peak is $12$ feet high. Each peak forms a $90^\circ$ angle, and the straight sides form a $45^\circ$ angle with the ground. The artwork has an area of $183$ square feet. The sides of the mountain meet at an intersection point near the center of the artwork, $h$ feet above the ground. What is the value of $h?$216[asy] unitsize(.3cm); filldraw((0,0)--(8,8)--(11,5)--(18,12)--(30,0)--cycle,gray(0.7),linewidth(1)); draw((-1,0)--(-1,8),linewidth(.75)); draw((-1.4,0)--(-.6,0),linewidth(.75)); draw((-1.4,8)--(-.6,8),linewidth(.75)); label(""$8$"",(-1,4),W); label(""$12$"",(31,6),E); draw((-1,8)--(8,8),dashed); draw((31,0)--(31,12),linewidth(.75)); draw((30.6,0)--(31.4,0),linewidth(.75)); draw((30.6,12)--(31.4,12),linewidth(.75)); draw((31,12)--(18,12),dashed); label(""$45^{\circ}$"",(.75,0),NE,fontsize(10pt)); label(""$45^{\circ}$"",(29.25,0),NW,fontsize(10pt)); draw((8,8)--(7.5,7.5)--(8,7)--(8.5,7.5)--cycle); draw((18,12)--(17.5,11.5)--(18,11)--(18.5,11.5)--cycle); draw((11,5)--(11,0),dashed); label(""$h$"",(11,2.5),E); [/asy]217$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 4\sqrt{2} \qquad \textbf{(D)}\ 5\sqrt{2} \qquad \textbf{(E)}\ 6$","You can measure $h$ with a ruler(rulers are allowed on the AMC 8), and see that $h$ is closest to $\boxed{5}.$",521853,https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_1,2,"What is the value of $(8 \times 4 + 2) - (8 + 4 \times 2)$219$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 24$",We can simplify the expression above in another way: \[(8 \times 4 + 2) - (8 + 4 \times 2)=8\times4+2-8-4\times2=32+2-8-8=34-16=\boxed{18}.\],1822054,https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_3,2,"Wind chill is a measure of how cold people feel when exposed to wind outside. A good estimate for wind chill can be found using this calculation \[(\text{wind chill}) = (\text{air temperature}) - 0.7 \times (\text{wind speed}),\] where temperature is measured in degrees Fahrenheit $(^{\circ}\text{F})$ and the wind speed is measured in miles per hour (mph). Suppose the air temperature is $36^{\circ}\text{F}$ and the wind speed is $18$ mph. Which of the following is closest to the approximate wind chill?221$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 23 \qquad \textbf{(C)}\ 28 \qquad \textbf{(D)}\ 32 \qquad \textbf{(E)}\ 35$","$0.7$ is very close to $\frac{2}{3}$ - therefore, we can substitute $\frac{2}{3}$ into the equation to get $36 - \frac{2}{3} * 18$ , which is $36 - 12 = 24$ . As $\frac{2}{3}$ is slightly less than $0.7$ , the correct answer is slightly less than $24$ . Therefore, the answer is $\boxed{23}$",2322255,https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_5,1,"A lake contains $250$ trout, along with a variety of other fish. When a marine biologist catches and releases a sample of $180$ fish from the lake, $30$ are identified as trout. Assume that the ratio of trout to the total number of fish is the same in both the sample and the lake. How many fish are there in the lake?223$\textbf{(A)}\ 1250 \qquad \textbf{(B)}\ 1500 \qquad \textbf{(C)}\ 1750 \qquad \textbf{(D)}\ 1800 \qquad \textbf{(E)}\ 2000$","Note that \[\frac{\text{number of trout}}{\text{total number of fish}} = \frac{30}{180} = \frac16.\] So, the total number of fish is $6$ times the number of trout. Since the lake contains $250$ trout, there are $250\cdot6=\boxed{1500}$ fish in the lake.",50022456,https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_6,1,"The digits $2,0,2,$ and $3$ are placed in the expression below, one digit per box. What is the maximum possible value of the expression?225[asy] // Diagram by TheMathGuyd. I can compress this later size(5cm); real w=2.2; pair O,I,J; O=(0,0);I=(1,0);J=(0,1); path bsqb = O--I; path bsqr = I--I+J; path bsqt = I+J--J; path bsql = J--O; path lsqb = shift((1.2,0.75))*scale(0.5)*bsqb; path lsqr = shift((1.2,0.75))*scale(0.5)*bsqr; path lsqt = shift((1.2,0.75))*scale(0.5)*bsqt; path lsql = shift((1.2,0.75))*scale(0.5)*bsql; draw(bsqb,dashed); draw(bsqr,dashed); draw(bsqt,dashed); draw(bsql,dashed); draw(lsqb,dashed); draw(lsqr,dashed); draw(lsqt,dashed); draw(lsql,dashed); label(scale(3)*""$\times$"",(w,1/3)); draw(shift(1.3w,0)*bsqb,dashed); draw(shift(1.3w,0)*bsqr,dashed); draw(shift(1.3w,0)*bsqt,dashed); draw(shift(1.3w,0)*bsql,dashed); draw(shift(1.3w,0)*lsqb,dashed); draw(shift(1.3w,0)*lsqr,dashed); draw(shift(1.3w,0)*lsqt,dashed); draw(shift(1.3w,0)*lsql,dashed); [/asy]226$\textbf{(A) }0 \qquad \textbf{(B) }8 \qquad \textbf{(C) }9 \qquad \textbf{(D) }16 \qquad \textbf{(E) }18$","First, let us consider the case where $0$ is a base: This would result in the entire expression being $0.$ Contrastingly, if $0$ is an exponent, we will get a value greater than $0.$ $3^2\times2^0=9$ is greater than $2^3\times2^0=8$ and $2^2\times3^0=4.$ Therefore, the answer is $\boxed{9}.$",922757,https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_6,2,"The digits $2,0,2,$ and $3$ are placed in the expression below, one digit per box. What is the maximum possible value of the expression?228[asy] // Diagram by TheMathGuyd. I can compress this later size(5cm); real w=2.2; pair O,I,J; O=(0,0);I=(1,0);J=(0,1); path bsqb = O--I; path bsqr = I--I+J; path bsqt = I+J--J; path bsql = J--O; path lsqb = shift((1.2,0.75))*scale(0.5)*bsqb; path lsqr = shift((1.2,0.75))*scale(0.5)*bsqr; path lsqt = shift((1.2,0.75))*scale(0.5)*bsqt; path lsql = shift((1.2,0.75))*scale(0.5)*bsql; draw(bsqb,dashed); draw(bsqr,dashed); draw(bsqt,dashed); draw(bsql,dashed); draw(lsqb,dashed); draw(lsqr,dashed); draw(lsqt,dashed); draw(lsql,dashed); label(scale(3)*""$\times$"",(w,1/3)); draw(shift(1.3w,0)*bsqb,dashed); draw(shift(1.3w,0)*bsqr,dashed); draw(shift(1.3w,0)*bsqt,dashed); draw(shift(1.3w,0)*bsql,dashed); draw(shift(1.3w,0)*lsqb,dashed); draw(shift(1.3w,0)*lsqr,dashed); draw(shift(1.3w,0)*lsqt,dashed); draw(shift(1.3w,0)*lsql,dashed); [/asy]229$\textbf{(A) }0 \qquad \textbf{(B) }8 \qquad \textbf{(C) }9 \qquad \textbf{(D) }16 \qquad \textbf{(E) }18$","The maximum possible value of using the digits $2,0,2,$ and $3$ : We can maximize our value by keeping the $3$ and $2$ together in one power (the biggest with the biggest and the smallest with the smallest). This shows $3^{2}\times2^{0}=9\times1=9.$ (We don't want $0^{2}$ because that is $0$ .) It is going to be $\boxed{9}.$",923058,https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_6,3,"The digits $2,0,2,$ and $3$ are placed in the expression below, one digit per box. What is the maximum possible value of the expression?231[asy] // Diagram by TheMathGuyd. I can compress this later size(5cm); real w=2.2; pair O,I,J; O=(0,0);I=(1,0);J=(0,1); path bsqb = O--I; path bsqr = I--I+J; path bsqt = I+J--J; path bsql = J--O; path lsqb = shift((1.2,0.75))*scale(0.5)*bsqb; path lsqr = shift((1.2,0.75))*scale(0.5)*bsqr; path lsqt = shift((1.2,0.75))*scale(0.5)*bsqt; path lsql = shift((1.2,0.75))*scale(0.5)*bsql; draw(bsqb,dashed); draw(bsqr,dashed); draw(bsqt,dashed); draw(bsql,dashed); draw(lsqb,dashed); draw(lsqr,dashed); draw(lsqt,dashed); draw(lsql,dashed); label(scale(3)*""$\times$"",(w,1/3)); draw(shift(1.3w,0)*bsqb,dashed); draw(shift(1.3w,0)*bsqr,dashed); draw(shift(1.3w,0)*bsqt,dashed); draw(shift(1.3w,0)*bsql,dashed); draw(shift(1.3w,0)*lsqb,dashed); draw(shift(1.3w,0)*lsqr,dashed); draw(shift(1.3w,0)*lsqt,dashed); draw(shift(1.3w,0)*lsql,dashed); [/asy]232$\textbf{(A) }0 \qquad \textbf{(B) }8 \qquad \textbf{(C) }9 \qquad \textbf{(D) }16 \qquad \textbf{(E) }18$","Trying all $12$ distinct orderings, we see that the only possible values are $0,4,8,$ and $9,$ the greatest of which is $\boxed{9}.$",923359,https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_7,1,"A rectangle, with sides parallel to the $x$ -axis and $y$ -axis, has opposite vertices located at $(15, 3)$ and $(16, 5)$ . A line is drawn through points $A(0, 0)$ and $B(3, 1)$ . Another line is drawn through points $C(0, 10)$ and $D(2, 9)$ . How many points on the rectangle lie on at least one of the two lines? [asy] usepackage(""mathptmx""); size(9cm); draw((0,-.5)--(0,11),EndArrow(size=.15cm)); draw((1,0)--(1,11),mediumgray); draw((2,0)--(2,11),mediumgray); draw((3,0)--(3,11),mediumgray); draw((4,0)--(4,11),mediumgray); draw((5,0)--(5,11),mediumgray); draw((6,0)--(6,11),mediumgray); draw((7,0)--(7,11),mediumgray); draw((8,0)--(8,11),mediumgray); draw((9,0)--(9,11),mediumgray); draw((10,0)--(10,11),mediumgray); draw((11,0)--(11,11),mediumgray); draw((12,0)--(12,11),mediumgray); draw((13,0)--(13,11),mediumgray); draw((14,0)--(14,11),mediumgray); draw((15,0)--(15,11),mediumgray); draw((16,0)--(16,11),mediumgray);  draw((-.5,0)--(17,0),EndArrow(size=.15cm)); draw((0,1)--(17,1),mediumgray); draw((0,2)--(17,2),mediumgray); draw((0,3)--(17,3),mediumgray); draw((0,4)--(17,4),mediumgray); draw((0,5)--(17,5),mediumgray); draw((0,6)--(17,6),mediumgray); draw((0,7)--(17,7),mediumgray); draw((0,8)--(17,8),mediumgray); draw((0,9)--(17,9),mediumgray); draw((0,10)--(17,10),mediumgray);  draw((-.13,1)--(.13,1)); draw((-.13,2)--(.13,2)); draw((-.13,3)--(.13,3)); draw((-.13,4)--(.13,4)); draw((-.13,5)--(.13,5)); draw((-.13,6)--(.13,6)); draw((-.13,7)--(.13,7)); draw((-.13,8)--(.13,8)); draw((-.13,9)--(.13,9)); draw((-.13,10)--(.13,10));  draw((1,-.13)--(1,.13)); draw((2,-.13)--(2,.13)); draw((3,-.13)--(3,.13)); draw((4,-.13)--(4,.13)); draw((5,-.13)--(5,.13)); draw((6,-.13)--(6,.13)); draw((7,-.13)--(7,.13)); draw((8,-.13)--(8,.13)); draw((9,-.13)--(9,.13)); draw((10,-.13)--(10,.13)); draw((11,-.13)--(11,.13)); draw((12,-.13)--(12,.13)); draw((13,-.13)--(13,.13)); draw((14,-.13)--(14,.13)); draw((15,-.13)--(15,.13)); draw((16,-.13)--(16,.13));  label(scale(.7)*""$1$"", (1,-.13), S); label(scale(.7)*""$2$"", (2,-.13), S); label(scale(.7)*""$3$"", (3,-.13), S); label(scale(.7)*""$4$"", (4,-.13), S); label(scale(.7)*""$5$"", (5,-.13), S); label(scale(.7)*""$6$"", (6,-.13), S); label(scale(.7)*""$7$"", (7,-.13), S); label(scale(.7)*""$8$"", (8,-.13), S); label(scale(.7)*""$9$"", (9,-.13), S); label(scale(.7)*""$10$"", (10,-.13), S); label(scale(.7)*""$11$"", (11,-.13), S); label(scale(.7)*""$12$"", (12,-.13), S); label(scale(.7)*""$13$"", (13,-.13), S); label(scale(.7)*""$14$"", (14,-.13), S); label(scale(.7)*""$15$"", (15,-.13), S); label(scale(.7)*""$16$"", (16,-.13), S);  label(scale(.7)*""$1$"", (-.13,1), W); label(scale(.7)*""$2$"", (-.13,2), W); label(scale(.7)*""$3$"", (-.13,3), W); label(scale(.7)*""$4$"", (-.13,4), W); label(scale(.7)*""$5$"", (-.13,5), W); label(scale(.7)*""$6$"", (-.13,6), W); label(scale(.7)*""$7$"", (-.13,7), W); label(scale(.7)*""$8$"", (-.13,8), W); label(scale(.7)*""$9$"", (-.13,9), W); label(scale(.7)*""$10$"", (-.13,10), W);   dot((0,0),linewidth(4)); label(scale(.75)*""$A$"", (0,0), NE); dot((3,1),linewidth(4)); label(scale(.75)*""$B$"", (3,1), NE);  dot((0,10),linewidth(4)); label(scale(.75)*""$C$"", (0,10), NE); dot((2,9),linewidth(4)); label(scale(.75)*""$D$"", (2,9), NE);  draw((15,3)--(16,3)--(16,5)--(15,5)--cycle,linewidth(1.125)); dot((15,3),linewidth(4)); dot((16,3),linewidth(4)); dot((16,5),linewidth(4)); dot((15,5),linewidth(4)); [/asy] $\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$","If we extend the lines, we have the following diagram: [asy] usepackage(""mathptmx""); size(9cm); draw((0,-.5)--(0,11),EndArrow(size=.15cm)); draw((1,0)--(1,11),mediumgray); draw((2,0)--(2,11),mediumgray); draw((3,0)--(3,11),mediumgray); draw((4,0)--(4,11),mediumgray); draw((5,0)--(5,11),mediumgray); draw((6,0)--(6,11),mediumgray); draw((7,0)--(7,11),mediumgray); draw((8,0)--(8,11),mediumgray); draw((9,0)--(9,11),mediumgray); draw((10,0)--(10,11),mediumgray); draw((11,0)--(11,11),mediumgray); draw((12,0)--(12,11),mediumgray); draw((13,0)--(13,11),mediumgray); draw((14,0)--(14,11),mediumgray); draw((15,0)--(15,11),mediumgray); draw((16,0)--(16,11),mediumgray);  draw((-.5,0)--(17,0),EndArrow(size=.15cm)); draw((0,1)--(17,1),mediumgray); draw((0,2)--(17,2),mediumgray); draw((0,3)--(17,3),mediumgray); draw((0,4)--(17,4),mediumgray); draw((0,5)--(17,5),mediumgray); draw((0,6)--(17,6),mediumgray); draw((0,7)--(17,7),mediumgray); draw((0,8)--(17,8),mediumgray); draw((0,9)--(17,9),mediumgray); draw((0,10)--(17,10),mediumgray);  draw((-.13,1)--(.13,1)); draw((-.13,2)--(.13,2)); draw((-.13,3)--(.13,3)); draw((-.13,4)--(.13,4)); draw((-.13,5)--(.13,5)); draw((-.13,6)--(.13,6)); draw((-.13,7)--(.13,7)); draw((-.13,8)--(.13,8)); draw((-.13,9)--(.13,9)); draw((-.13,10)--(.13,10));  draw((1,-.13)--(1,.13)); draw((2,-.13)--(2,.13)); draw((3,-.13)--(3,.13)); draw((4,-.13)--(4,.13)); draw((5,-.13)--(5,.13)); draw((6,-.13)--(6,.13)); draw((7,-.13)--(7,.13)); draw((8,-.13)--(8,.13)); draw((9,-.13)--(9,.13)); draw((10,-.13)--(10,.13)); draw((11,-.13)--(11,.13)); draw((12,-.13)--(12,.13)); draw((13,-.13)--(13,.13)); draw((14,-.13)--(14,.13)); draw((15,-.13)--(15,.13)); draw((16,-.13)--(16,.13));  label(scale(.7)*""$1$"", (1,-.13), S); label(scale(.7)*""$2$"", (2,-.13), S); label(scale(.7)*""$3$"", (3,-.13), S); label(scale(.7)*""$4$"", (4,-.13), S); label(scale(.7)*""$5$"", (5,-.13), S); label(scale(.7)*""$6$"", (6,-.13), S); label(scale(.7)*""$7$"", (7,-.13), S); label(scale(.7)*""$8$"", (8,-.13), S); label(scale(.7)*""$9$"", (9,-.13), S); label(scale(.7)*""$10$"", (10,-.13), S); label(scale(.7)*""$11$"", (11,-.13), S); label(scale(.7)*""$12$"", (12,-.13), S); label(scale(.7)*""$13$"", (13,-.13), S); label(scale(.7)*""$14$"", (14,-.13), S); label(scale(.7)*""$15$"", (15,-.13), S); label(scale(.7)*""$16$"", (16,-.13), S);  label(scale(.7)*""$1$"", (-.13,1), W); label(scale(.7)*""$2$"", (-.13,2), W); label(scale(.7)*""$3$"", (-.13,3), W); label(scale(.7)*""$4$"", (-.13,4), W); label(scale(.7)*""$5$"", (-.13,5), W); label(scale(.7)*""$6$"", (-.13,6), W); label(scale(.7)*""$7$"", (-.13,7), W); label(scale(.7)*""$8$"", (-.13,8), W); label(scale(.7)*""$9$"", (-.13,9), W); label(scale(.7)*""$10$"", (-.13,10), W);  draw((0,10)--(17,1.5),blue); draw((0,0)--(17,17/3),blue);  dot((0,0),linewidth(4)); label(scale(.75)*""$A$"", (0,0), NE); dot((3,1),linewidth(4)); label(scale(.75)*""$B$"", (3,1), NE);  dot((0,10),linewidth(4)); label(scale(.75)*""$C$"", (0,10), NE); dot((2,9),linewidth(4)); label(scale(.75)*""$D$"", (2,9), NE);  draw((15,3)--(16,3)--(16,5)--(15,5)--cycle,linewidth(1.125)); dot((15,3),linewidth(4)); dot((16,3),linewidth(4)); dot((16,5),linewidth(4)); dot((15,5),linewidth(4)); [/asy] Therefore, we see that the answer is $\boxed{1}.$",123460,https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_7,2,"A rectangle, with sides parallel to the $x$ -axis and $y$ -axis, has opposite vertices located at $(15, 3)$ and $(16, 5)$ . A line is drawn through points $A(0, 0)$ and $B(3, 1)$ . Another line is drawn through points $C(0, 10)$ and $D(2, 9)$ . How many points on the rectangle lie on at least one of the two lines? [asy] usepackage(""mathptmx""); size(9cm); draw((0,-.5)--(0,11),EndArrow(size=.15cm)); draw((1,0)--(1,11),mediumgray); draw((2,0)--(2,11),mediumgray); draw((3,0)--(3,11),mediumgray); draw((4,0)--(4,11),mediumgray); draw((5,0)--(5,11),mediumgray); draw((6,0)--(6,11),mediumgray); draw((7,0)--(7,11),mediumgray); draw((8,0)--(8,11),mediumgray); draw((9,0)--(9,11),mediumgray); draw((10,0)--(10,11),mediumgray); draw((11,0)--(11,11),mediumgray); draw((12,0)--(12,11),mediumgray); draw((13,0)--(13,11),mediumgray); draw((14,0)--(14,11),mediumgray); draw((15,0)--(15,11),mediumgray); draw((16,0)--(16,11),mediumgray);  draw((-.5,0)--(17,0),EndArrow(size=.15cm)); draw((0,1)--(17,1),mediumgray); draw((0,2)--(17,2),mediumgray); draw((0,3)--(17,3),mediumgray); draw((0,4)--(17,4),mediumgray); draw((0,5)--(17,5),mediumgray); draw((0,6)--(17,6),mediumgray); draw((0,7)--(17,7),mediumgray); draw((0,8)--(17,8),mediumgray); draw((0,9)--(17,9),mediumgray); draw((0,10)--(17,10),mediumgray);  draw((-.13,1)--(.13,1)); draw((-.13,2)--(.13,2)); draw((-.13,3)--(.13,3)); draw((-.13,4)--(.13,4)); draw((-.13,5)--(.13,5)); draw((-.13,6)--(.13,6)); draw((-.13,7)--(.13,7)); draw((-.13,8)--(.13,8)); draw((-.13,9)--(.13,9)); draw((-.13,10)--(.13,10));  draw((1,-.13)--(1,.13)); draw((2,-.13)--(2,.13)); draw((3,-.13)--(3,.13)); draw((4,-.13)--(4,.13)); draw((5,-.13)--(5,.13)); draw((6,-.13)--(6,.13)); draw((7,-.13)--(7,.13)); draw((8,-.13)--(8,.13)); draw((9,-.13)--(9,.13)); draw((10,-.13)--(10,.13)); draw((11,-.13)--(11,.13)); draw((12,-.13)--(12,.13)); draw((13,-.13)--(13,.13)); draw((14,-.13)--(14,.13)); draw((15,-.13)--(15,.13)); draw((16,-.13)--(16,.13));  label(scale(.7)*""$1$"", (1,-.13), S); label(scale(.7)*""$2$"", (2,-.13), S); label(scale(.7)*""$3$"", (3,-.13), S); label(scale(.7)*""$4$"", (4,-.13), S); label(scale(.7)*""$5$"", (5,-.13), S); label(scale(.7)*""$6$"", (6,-.13), S); label(scale(.7)*""$7$"", (7,-.13), S); label(scale(.7)*""$8$"", (8,-.13), S); label(scale(.7)*""$9$"", (9,-.13), S); label(scale(.7)*""$10$"", (10,-.13), S); label(scale(.7)*""$11$"", (11,-.13), S); label(scale(.7)*""$12$"", (12,-.13), S); label(scale(.7)*""$13$"", (13,-.13), S); label(scale(.7)*""$14$"", (14,-.13), S); label(scale(.7)*""$15$"", (15,-.13), S); label(scale(.7)*""$16$"", (16,-.13), S);  label(scale(.7)*""$1$"", (-.13,1), W); label(scale(.7)*""$2$"", (-.13,2), W); label(scale(.7)*""$3$"", (-.13,3), W); label(scale(.7)*""$4$"", (-.13,4), W); label(scale(.7)*""$5$"", (-.13,5), W); label(scale(.7)*""$6$"", (-.13,6), W); label(scale(.7)*""$7$"", (-.13,7), W); label(scale(.7)*""$8$"", (-.13,8), W); label(scale(.7)*""$9$"", (-.13,9), W); label(scale(.7)*""$10$"", (-.13,10), W);   dot((0,0),linewidth(4)); label(scale(.75)*""$A$"", (0,0), NE); dot((3,1),linewidth(4)); label(scale(.75)*""$B$"", (3,1), NE);  dot((0,10),linewidth(4)); label(scale(.75)*""$C$"", (0,10), NE); dot((2,9),linewidth(4)); label(scale(.75)*""$D$"", (2,9), NE);  draw((15,3)--(16,3)--(16,5)--(15,5)--cycle,linewidth(1.125)); dot((15,3),linewidth(4)); dot((16,3),linewidth(4)); dot((16,5),linewidth(4)); dot((15,5),linewidth(4)); [/asy] $\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$","Note that the $y$ -intercepts of line $AB$ and line $CD$ are $0$ and $10$ . If the analytic expression for line $AB$ is $y=k_{1}x$ , and the analytic expression for line $CD$ is $y=k_{2}x+10$ , we have equations: $3k_{1} = 1$ and $2k_{2} + 10 = 9$ . Solving these equations, we can find out that $k_{1} = \frac{1}{3}$ and $k_{2} = -\frac{1}{2}$ . Therefore, we can determine that the expression for line $AB$ is $y=\frac{1}{3}x$ and the expression for line $CD$ is $y=-\frac{1}{2}x + 10$ . When $x=15$ , the coordinates that line $AB$ and line $CD$ pass through are $(15, 5)$ and $\left(15, \frac{5}{2}\right)$ , and $(15, 5)$ lies perfectly on one vertex of the rectangle while the $y$ coordinate of $\left(15, \frac{5}{2}\right)$ is out of the range $3 \leq y \leq 5$ (lower than the bottom left corner of the rectangle $(15, 3)$ ). Considering that the $y$ value of the line $CD$ will only decrease, and the $y$ value of the line $AB$ will only increase, there will not be another point on the rectangle that lies on either of the two lines. Thus, we can conclude that the answer is $\boxed{1}.$",123561,https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_9,1,"Malaika is skiing on a mountain. The graph below shows her elevation, in meters, above the base of the mountain as she skis along a trail. In total, how many seconds does she spend at an elevation between $4$ and $7$ meters? [asy] // Diagram by TheMathGuyd. Found cubic, so graph is perfect. import graph; size(8cm); int i; for(i=1; i<9; i=i+1) { draw((-0.2,2i-1)--(16.2,2i-1), mediumgrey); draw((2i-1,-0.2)--(2i-1,16.2), mediumgrey); draw((-0.2,2i)--(16.2,2i), grey); draw((2i,-0.2)--(2i,16.2), grey); } Label f;  f.p=fontsize(6);  xaxis(-0.5,17.8,Ticks(f, 2.0),Arrow());  yaxis(-0.5,17.8,Ticks(f, 2.0),Arrow());  real f(real x)  {  return -0.03125 x^(3) + 0.75x^(2) - 5.125 x + 14.5;  }  draw(graph(f,0,15.225),currentpen+1); real dpt=2; real ts=0.75; transform st=scale(ts); label(rotate(90)*st*""Elevation (meters)"",(-dpt,8)); label(st*""Time (seconds)"",(8,-dpt)); [/asy] $\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 14$","We mark the time intervals in which Malaika's elevation is between $4$ and $7$ meters in red, as shown below: [asy] // Diagram by TheMathGuyd. Found cubic, so graph is perfect. import graph; size(8cm); int i; for(i=1; i<9; i=i+1) { draw((-0.2,2i-1)--(16.2,2i-1), mediumgrey); draw((2i-1,-0.2)--(2i-1,16.2), mediumgrey); draw((-0.2,2i)--(16.2,2i), grey); draw((2i,-0.2)--(2i,16.2), grey); } Label f;  f.p=fontsize(6);  xaxis(-0.5,17.8,Ticks(f, 2.0),Arrow());  yaxis(-0.5,17.8,Ticks(f, 2.0),Arrow());  real f(real x)  {  return -0.03125 x^(3) + 0.75x^(2) - 5.125 x + 14.5;  }  draw(graph(f,0,15.225),currentpen+1); draw(graph(f,2,4)^^graph(f,6,10)^^graph(f,12,14),red+currentpen+2); real dpt=2; real ts=0.75; transform st=scale(ts); label(rotate(90)*st*""Elevation (meters)"",(-dpt,8)); label(st*""Time (seconds)"",(8,-dpt)); [/asy] The requested time intervals are:236In total, Malaika spends $(4-2) + (10-6) + (14-12) = \boxed{8}$ seconds at such elevation.",823762,https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_9,2,"Malaika is skiing on a mountain. The graph below shows her elevation, in meters, above the base of the mountain as she skis along a trail. In total, how many seconds does she spend at an elevation between $4$ and $7$ meters? [asy] // Diagram by TheMathGuyd. Found cubic, so graph is perfect. import graph; size(8cm); int i; for(i=1; i<9; i=i+1) { draw((-0.2,2i-1)--(16.2,2i-1), mediumgrey); draw((2i-1,-0.2)--(2i-1,16.2), mediumgrey); draw((-0.2,2i)--(16.2,2i), grey); draw((2i,-0.2)--(2i,16.2), grey); } Label f;  f.p=fontsize(6);  xaxis(-0.5,17.8,Ticks(f, 2.0),Arrow());  yaxis(-0.5,17.8,Ticks(f, 2.0),Arrow());  real f(real x)  {  return -0.03125 x^(3) + 0.75x^(2) - 5.125 x + 14.5;  }  draw(graph(f,0,15.225),currentpen+1); real dpt=2; real ts=0.75; transform st=scale(ts); label(rotate(90)*st*""Elevation (meters)"",(-dpt,8)); label(st*""Time (seconds)"",(8,-dpt)); [/asy] $\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 14$","Notice that the entire section between the $2$ second mark and the $14$ second mark is between the $4$ and $7$ feet elevation level except the $2$ seconds where she skis just under the $4$ feet mark and when she skis just above the $7$ feet mark, making the answer $14-2-2-2=\boxed{8}.$",823863,https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_11,1,"NASA’s Perseverance Rover was launched on July $30,$ $2020.$ After traveling $292526838$ miles, it landed on Mars in Jezero Crater about $6.5$ months later. Which of the following is closest to the Rover’s average interplanetary speed in miles per hour?239$\textbf{(A)}\ 6000 \qquad \textbf{(B)}\ 12000 \qquad \textbf{(C)}\ 60000 \qquad \textbf{(D)}\ 120000 \qquad \textbf{(E)}\ 600000$","Note that $6.5$ months is approximately $6.5\cdot30\cdot24$ hours. Therefore, the speed (in miles per hour) is \[\frac{292526838}{6.5\cdot30\cdot24} \approx \frac{300000000}{6.5\cdot30\cdot24} = \frac{10000000}{6.5\cdot24} \approx \frac{10000000}{6.4\cdot25} = \frac{10000000}{160} = 62500 \approx \boxed{60000}.\] As the answer choices are far apart from each other, we can ensure that the approximation is correct.",024064,https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_11,2,"NASA’s Perseverance Rover was launched on July $30,$ $2020.$ After traveling $292526838$ miles, it landed on Mars in Jezero Crater about $6.5$ months later. Which of the following is closest to the Rover’s average interplanetary speed in miles per hour?241$\textbf{(A)}\ 6000 \qquad \textbf{(B)}\ 12000 \qquad \textbf{(C)}\ 60000 \qquad \textbf{(D)}\ 120000 \qquad \textbf{(E)}\ 600000$","Note that $292526838 \approx 300000000$ miles. We also know that $6.5$ months is approximately $6.5\cdot30\cdot24$ hours. Now, we can calculate the speed in miles per hour, which we find is about \[\dfrac{300000000}{6.5\cdot30\cdot24}=\dfrac{10000000}{6.5\cdot24}=\dfrac{10000000}{13\cdot12}=\dfrac{10000000}{156}\approx\dfrac{10000000}{150}\approx\dfrac{200000}{3}\approx\boxed{60000}.\] ~MathFun1000",024265,https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_14,1,"Nicolas is planning to send a package to his friend Anton, who is a stamp collector. To pay for the postage, Nicolas would like to cover the package with a large number of stamps. Suppose he has a collection of $5$ -cent, $10$ -cent, and $25$ -cent stamps, with exactly $20$ of each type. What is the greatest number of stamps Nicolas can use to make exactly $$7.10$ in postage?243(Note: The amount $$7.10$ corresponds to $7$ dollars and $10$ cents. One dollar is worth $100$ cents.)244$\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 46 \qquad \textbf{(C)}\ 51 \qquad \textbf{(D)}\ 54\qquad \textbf{(E)}\ 55$","Let's use the most stamps to make $7.10.$ We have $20$ of each stamp, $5$ -cent (nickels), $10$ -cent (dimes), and $25$ -cent (quarters).245If we want the highest number of stamps, we must have the highest number of the smaller value stamps (like the coins above). We can use $20$ nickels and $20$ dimes to bring our total cost to $7.10 - 3.00 = 4.10$ . However, when we try to use quarters, the $25$ cents don’t fit evenly, so we have to give back $15$ cents to make the quarter amount $4.25$ . The most efficient way to do this is to give back a $10$ -cent (dime) stamp and a $5$ -cent (nickel) stamp to have $38$ stamps used so far. Now, we just use $\frac{425}{25} = 17$ quarters to get a grand total of $38 + 17 = \boxed{55}$",5524666,https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_14,2,"Nicolas is planning to send a package to his friend Anton, who is a stamp collector. To pay for the postage, Nicolas would like to cover the package with a large number of stamps. Suppose he has a collection of $5$ -cent, $10$ -cent, and $25$ -cent stamps, with exactly $20$ of each type. What is the greatest number of stamps Nicolas can use to make exactly $$7.10$ in postage?247(Note: The amount $$7.10$ corresponds to $7$ dollars and $10$ cents. One dollar is worth $100$ cents.)248$\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 46 \qquad \textbf{(C)}\ 51 \qquad \textbf{(D)}\ 54\qquad \textbf{(E)}\ 55$","The value of his entire stamp collection is $8$ dollars. To make $$7.10$ with stamps, he should remove $90$ cents worth of stamps with as few stamps as possible. To do this, he should start by removing as many $25$ cent stamps as possible as they have the greatest denomination. He can remove at most $3$ of these stamps. He still has to remove $90-25\cdot3=15$ cents worth of stamps. This can be done with one $5$ and $10$ cent stamp. In total, he has $20\cdot3=60$ stamps in his entire collection. As a result, the maximum number of stamps he can use is $20\cdot3-5=\boxed{55}$",5524967,https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_17,1,"regular octahedron has eight equilateral triangle faces with four faces meeting at each vertex. Jun will make the regular octahedron shown on the right by folding the piece of paper shown on the left. Which numbered face will end up to the right of $Q$250[asy] // Diagram by TheMathGuyd import graph; // The Solid // To save processing time, do not use three (dimensions) // Project (roughly) to two size(15cm); pair Fr, Lf, Rt, Tp, Bt, Bk; Lf=(0,0); Rt=(12,1); Fr=(7,-1); Bk=(5,2); Tp=(6,6.7); Bt=(6,-5.2); draw(Lf--Fr--Rt); draw(Lf--Tp--Rt); draw(Lf--Bt--Rt); draw(Tp--Fr--Bt); draw(Lf--Bk--Rt,dashed); draw(Tp--Bk--Bt,dashed); label(rotate(-8.13010235)*slant(0.1)*""$Q$"", (4.2,1.6)); label(rotate(21.8014095)*slant(-0.2)*""$?$"", (8.5,2.05)); pair g = (-8,0); // Define Gap transform real a = 8; draw(g+(-a/2,1)--g+(a/2,1), Arrow()); // Make arrow // Time for the NET pair DA,DB,DC,CD,O; DA = (4*sqrt(3),0); DB = (2*sqrt(3),6); DC = (DA+DB)/3; CD = conj(DC); O=(0,0); transform trf=shift(3g+(0,3)); path NET = O--(-2*DA)--(-2DB)--(-DB)--(2DA-DB)--DB--O--DA--(DA-DB)--O--(-DB)--(-DA)--(-DA-DB)--(-DB); draw(trf*NET); label(""$7$"",trf*DC); label(""$Q$"",trf*DC+DA-DB); label(""$5$"",trf*DC-DB); label(""$3$"",trf*DC-DA-DB); label(""$6$"",trf*CD); label(""$4$"",trf*CD-DA); label(""$2$"",trf*CD-DA-DB); label(""$1$"",trf*CD-2DA); [/asy]251$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$","We color face $6$ red and face $5$ yellow. Note that from the octahedron, face $5$ and face $?$ do not share anything in common. From the net, face $5$ shares at least one vertex with all other faces except face $1,$ which is shown in green: [asy] /* Diagram by TheMathGuyd Edited by MRENTHUSIASM */ import graph; // The Solid // To save processing time, do not use three (dimensions) // Project (roughly) to two size(15cm); pair Fr, Lf, Rt, Tp, Bt, Bk; Lf=(0,0); Rt=(12,1); Fr=(7,-1); Bk=(5,2); Tp=(6,6.7); Bt=(6,-5.2); fill(Tp--Bk--Lf--cycle,red); fill(Bt--Bk--Lf--cycle,yellow); fill(Fr--Rt--Tp--cycle,green); draw(Lf--Fr--Rt); draw(Lf--Tp--Rt); draw(Lf--Bt--Rt); draw(Tp--Fr--Bt); draw(Lf--Bk--Rt,dashed); draw(Tp--Bk--Bt,dashed); label(rotate(-8.13010235)*slant(0.1)*""$Q$"", (4.2,1.6)); label(rotate(21.8014095)*slant(-0.2)*""$?$"", (8.5,2.05)); pair g = (-8,0); // Define Gap transform real a = 8; draw(g+(-a/2,1)--g+(a/2,1), Arrow()); // Make arrow // Time for the NET pair DA,DB,DC,CD,O; DA = (4*sqrt(3),0); DB = (2*sqrt(3),6); DC = (DA+DB)/3; CD = conj(DC); O=(0,0); transform trf=shift(3g+(0,3)); path NET = O--(-2*DA)--(-2DB)--(-DB)--(2DA-DB)--DB--O--DA--(DA-DB)--O--(-DB)--(-DA)--(-DA-DB)--(-DB); fill(trf*((DA-DB)--O--DA--cycle),red); fill(trf*((DA-DB)--O--(-DB)--cycle),yellow); fill(trf*((-2*DA)--(-DA-DB)--(-DA)--cycle),green); draw(trf*NET); label(""$7$"",trf*DC); label(""$Q$"",trf*DC+DA-DB); label(""$5$"",trf*DC-DB); label(""$3$"",trf*DC-DA-DB); label(""$6$"",trf*CD); label(""$4$"",trf*CD-DA); label(""$2$"",trf*CD-DA-DB); label(""$1$"",trf*CD-2DA); [/asy] Therefore, the answer is $\boxed{1}.$",125268,https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_17,2,"regular octahedron has eight equilateral triangle faces with four faces meeting at each vertex. Jun will make the regular octahedron shown on the right by folding the piece of paper shown on the left. Which numbered face will end up to the right of $Q$253[asy] // Diagram by TheMathGuyd import graph; // The Solid // To save processing time, do not use three (dimensions) // Project (roughly) to two size(15cm); pair Fr, Lf, Rt, Tp, Bt, Bk; Lf=(0,0); Rt=(12,1); Fr=(7,-1); Bk=(5,2); Tp=(6,6.7); Bt=(6,-5.2); draw(Lf--Fr--Rt); draw(Lf--Tp--Rt); draw(Lf--Bt--Rt); draw(Tp--Fr--Bt); draw(Lf--Bk--Rt,dashed); draw(Tp--Bk--Bt,dashed); label(rotate(-8.13010235)*slant(0.1)*""$Q$"", (4.2,1.6)); label(rotate(21.8014095)*slant(-0.2)*""$?$"", (8.5,2.05)); pair g = (-8,0); // Define Gap transform real a = 8; draw(g+(-a/2,1)--g+(a/2,1), Arrow()); // Make arrow // Time for the NET pair DA,DB,DC,CD,O; DA = (4*sqrt(3),0); DB = (2*sqrt(3),6); DC = (DA+DB)/3; CD = conj(DC); O=(0,0); transform trf=shift(3g+(0,3)); path NET = O--(-2*DA)--(-2DB)--(-DB)--(2DA-DB)--DB--O--DA--(DA-DB)--O--(-DB)--(-DA)--(-DA-DB)--(-DB); draw(trf*NET); label(""$7$"",trf*DC); label(""$Q$"",trf*DC+DA-DB); label(""$5$"",trf*DC-DB); label(""$3$"",trf*DC-DA-DB); label(""$6$"",trf*CD); label(""$4$"",trf*CD-DA); label(""$2$"",trf*CD-DA-DB); label(""$1$"",trf*CD-2DA); [/asy]254$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$","We label the octohedron going triangle by triangle until we reach the $?$ triangle. The triangle to the left of the $Q$ should be labeled with a $6$ . Underneath triangle $6$ is triangle $5$ . The triangle to the right of triangle $5$ is triangle $4$ and further to the right is triangle $3$ . Finally, the side of triangle $3$ under triangle $Q$ is $2$ , so the triangle to the right of $Q$ is $\boxed{1}$",125569,https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_17,3,"regular octahedron has eight equilateral triangle faces with four faces meeting at each vertex. Jun will make the regular octahedron shown on the right by folding the piece of paper shown on the left. Which numbered face will end up to the right of $Q$256[asy] // Diagram by TheMathGuyd import graph; // The Solid // To save processing time, do not use three (dimensions) // Project (roughly) to two size(15cm); pair Fr, Lf, Rt, Tp, Bt, Bk; Lf=(0,0); Rt=(12,1); Fr=(7,-1); Bk=(5,2); Tp=(6,6.7); Bt=(6,-5.2); draw(Lf--Fr--Rt); draw(Lf--Tp--Rt); draw(Lf--Bt--Rt); draw(Tp--Fr--Bt); draw(Lf--Bk--Rt,dashed); draw(Tp--Bk--Bt,dashed); label(rotate(-8.13010235)*slant(0.1)*""$Q$"", (4.2,1.6)); label(rotate(21.8014095)*slant(-0.2)*""$?$"", (8.5,2.05)); pair g = (-8,0); // Define Gap transform real a = 8; draw(g+(-a/2,1)--g+(a/2,1), Arrow()); // Make arrow // Time for the NET pair DA,DB,DC,CD,O; DA = (4*sqrt(3),0); DB = (2*sqrt(3),6); DC = (DA+DB)/3; CD = conj(DC); O=(0,0); transform trf=shift(3g+(0,3)); path NET = O--(-2*DA)--(-2DB)--(-DB)--(2DA-DB)--DB--O--DA--(DA-DB)--O--(-DB)--(-DA)--(-DA-DB)--(-DB); draw(trf*NET); label(""$7$"",trf*DC); label(""$Q$"",trf*DC+DA-DB); label(""$5$"",trf*DC-DB); label(""$3$"",trf*DC-DA-DB); label(""$6$"",trf*CD); label(""$4$"",trf*CD-DA); label(""$2$"",trf*CD-DA-DB); label(""$1$"",trf*CD-2DA); [/asy]257$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$","Notice that the triangles labeled $2, 3, 4,$ and $5$ make the bottom half of the octahedron, as shown below: [asy] /* Diagram by TheMathGuyd Edited by MRENTHUSIASM */ import graph; // The Solid // To save processing time, do not use three (dimensions) // Project (roughly) to two size(15cm); pair Fr, Lf, Rt, Tp, Bt, Bk; Lf=(0,0); Rt=(12,1); Fr=(7,-1); Bk=(5,2); Tp=(6,6.7); Bt=(6,-5.2); dot(Bt,linewidth(5)); draw(Lf--Fr--Rt); draw(Lf--Tp--Rt); draw(Lf--Bt--Rt); draw(Tp--Fr--Bt); draw(Lf--Bk--Rt,dashed); draw(Tp--Bk--Bt,dashed); label(rotate(-8.13010235)*slant(0.1)*""$Q$"", (4.2,1.6)); label(rotate(21.8014095)*slant(-0.2)*""$?$"", (8.5,2.05)); pair g = (-8,0); // Define Gap transform real a = 8; draw(g+(-a/2,1)--g+(a/2,1), Arrow()); // Make arrow // Time for the NET pair DA,DB,DC,CD,O; DA = (4*sqrt(3),0); DB = (2*sqrt(3),6); DC = (DA+DB)/3; CD = conj(DC); O=(0,0); transform trf=shift(3g+(0,3)); path NET = O--(-2*DA)--(-2DB)--(-DB)--(2DA-DB)--DB--O--DA--(DA-DB)--O--(-DB)--(-DA)--(-DA-DB)--(-DB); dot(trf*(-DB),linewidth(5)); draw(trf*NET); label(""$7$"",trf*DC); label(""$Q$"",trf*DC+DA-DB); label(""$5$"",trf*DC-DB); label(""$3$"",trf*DC-DA-DB); label(""$6$"",trf*CD); label(""$4$"",trf*CD-DA); label(""$2$"",trf*CD-DA-DB); label(""$1$"",trf*CD-2DA); [/asy] Therefore, $\textbf{(B)}, \textbf{(C)}, \textbf{(D)},$ and $\textbf{(E)}$ are clearly not the correct answer. Thus, the only choice left is $\boxed{1}$",125870,https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_17,4,"regular octahedron has eight equilateral triangle faces with four faces meeting at each vertex. Jun will make the regular octahedron shown on the right by folding the piece of paper shown on the left. Which numbered face will end up to the right of $Q$259[asy] // Diagram by TheMathGuyd import graph; // The Solid // To save processing time, do not use three (dimensions) // Project (roughly) to two size(15cm); pair Fr, Lf, Rt, Tp, Bt, Bk; Lf=(0,0); Rt=(12,1); Fr=(7,-1); Bk=(5,2); Tp=(6,6.7); Bt=(6,-5.2); draw(Lf--Fr--Rt); draw(Lf--Tp--Rt); draw(Lf--Bt--Rt); draw(Tp--Fr--Bt); draw(Lf--Bk--Rt,dashed); draw(Tp--Bk--Bt,dashed); label(rotate(-8.13010235)*slant(0.1)*""$Q$"", (4.2,1.6)); label(rotate(21.8014095)*slant(-0.2)*""$?$"", (8.5,2.05)); pair g = (-8,0); // Define Gap transform real a = 8; draw(g+(-a/2,1)--g+(a/2,1), Arrow()); // Make arrow // Time for the NET pair DA,DB,DC,CD,O; DA = (4*sqrt(3),0); DB = (2*sqrt(3),6); DC = (DA+DB)/3; CD = conj(DC); O=(0,0); transform trf=shift(3g+(0,3)); path NET = O--(-2*DA)--(-2DB)--(-DB)--(2DA-DB)--DB--O--DA--(DA-DB)--O--(-DB)--(-DA)--(-DA-DB)--(-DB); draw(trf*NET); label(""$7$"",trf*DC); label(""$Q$"",trf*DC+DA-DB); label(""$5$"",trf*DC-DB); label(""$3$"",trf*DC-DA-DB); label(""$6$"",trf*CD); label(""$4$"",trf*CD-DA); label(""$2$"",trf*CD-DA-DB); label(""$1$"",trf*CD-2DA); [/asy]260$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$","The first half of the octahedron will need $4$ triangles connected to one another to form it. We can choose the triangles $4$ $5$ $6$ , and $7$ and form the half around the vertex they all share. That leaves triangles $1$ $3$ $2$ , and $Q$ to form the second half. Triangle $3$ will definitely share its sides with triangles $1$ and $2$ , leaving them to share their second side with triangle $Q$ . Since triangle $Q$ will certainly share its left side with triangle $2$ , the only triangle left to share its right side is triangle $\boxed{1}$",126171,https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_18,1,"Greta Grasshopper sits on a long line of lily pads in a pond. From any lily pad, Greta can jump $5$ pads to the right or $3$ pads to the left. What is the fewest number of jumps Greta must make to reach the lily pad located $2023$ pads to the right of her starting position?262$\textbf{(A) } 405 \qquad \textbf{(B) } 407 \qquad \textbf{(C) } 409 \qquad \textbf{(D) } 411 \qquad \textbf{(E) } 413$","We have $2$ directions going $5$ right or $3$ left. We can assign a variable to each of these directions. We can call going right $1$ direction $\text{X}$ and we can call going $1$ left $\text{Y}$ . We can build a equation of $5\text{X}-3\text{Y}=2023$ , where we have to limit the number of moves we do. We can do this by making more of our moves the $5$ move turn then the $3$ move turn. The first obvious step is to go some amount of moves in the right direction then subtract off in the left direction to land on $2023$ . The least amount of $3$ ’s added to $2023$ to make a multiple of $5$ is $4$ as $2023 + 4(3) = 2035$ . So now, we have solved the problem as we just go $\frac{2035}{5} = 407$ hops right, and just do 4 more hops left. Yielding $407 + 4 = \boxed{411}$ as our answer.",41126372,https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_18,2,"Greta Grasshopper sits on a long line of lily pads in a pond. From any lily pad, Greta can jump $5$ pads to the right or $3$ pads to the left. What is the fewest number of jumps Greta must make to reach the lily pad located $2023$ pads to the right of her starting position?264$\textbf{(A) } 405 \qquad \textbf{(B) } 407 \qquad \textbf{(C) } 409 \qquad \textbf{(D) } 411 \qquad \textbf{(E) } 413$","Notice that $2023 \equiv 3\pmod{5}$ , and jumping to the left increases the value of Greta's position $\pmod{5}$ by $2$ . Therefore, the number of jumps to the left must be $4 \pmod{5}$ . As the number of jumps to the left increases, so does the number of jumps to the right, we must minimize both, which occurs when we jump $4$ to the left and $407$ to the right. The answer is $\boxed{411}$",41126573,https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_18,3,"Greta Grasshopper sits on a long line of lily pads in a pond. From any lily pad, Greta can jump $5$ pads to the right or $3$ pads to the left. What is the fewest number of jumps Greta must make to reach the lily pad located $2023$ pads to the right of her starting position?266$\textbf{(A) } 405 \qquad \textbf{(B) } 407 \qquad \textbf{(C) } 409 \qquad \textbf{(D) } 411 \qquad \textbf{(E) } 413$","$5y - 2023$ must be divisible by 3. The smallest value of $y$ that will achieve this is $407$ , which lands it at $2035$ . After that, it takes $4$ jumps back, making a total of $\boxed{411}$",41126774,https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_20,1,"Two integers are inserted into the list $3, 3, 8, 11, 28$ to double its range. The mode and median remain unchanged. What is the maximum possible sum of the two additional numbers?268$\textbf{(A) } 56 \qquad \textbf{(B) } 57 \qquad \textbf{(C) } 58 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 61$","To double the range, we must find the current range, which is $28 - 3 = 25$ , to then double to: $2(25) = 50$ . Since we do not want to change the median, we need to get a value less than $8$ (as $8$ would change the mode) for the smaller, making $53$ fixed for the larger. Remember, anything less than $3$ is not beneficial to the optimization because you want to get the largest range without changing the mode. So, taking our optimal values of $7$ and $53$ , we have an answer of $7 + 53 = \boxed{60}$",6026975,https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_21,2,"Alina writes the numbers $1, 2, \dots , 9$ on separate cards, one number per card. She wishes to divide the cards into $3$ groups of $3$ cards so that the sum of the numbers in each group will be the same. In how many ways can this be done?270$\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4$","The group with $5$ must have the two other numbers adding up to $10$ , since the sum of all the numbers is $(1 + 2 \cdots + 9) = \frac{9(10)}{2} = 45$ . The sum of the numbers in each group must therefore be $\frac{45}{3}=15$ . We can have $(1, 5, 9)$ $(2, 5, 8)$ $(3, 5, 7)$ , or $(4, 5, 6)$ . With the first group, we have $(2, 3, 4, 6, 7, 8)$ left over. The only way to form a group of $3$ numbers that add up to $15$ is with $(3, 4, 8)$ or $(2, 6, 7)$ . One of the possible arrangements is therefore $(1, 5, 9) (3, 4, 8) (2, 6, 7)$ . Then, with the second group, we have $(1, 3, 4, 6, 7, 9)$ left over. With these numbers, there is no way to form a group of $3$ numbers adding to $15$ . Similarly, with the third group there is $(1, 2, 4, 6, 8, 9)$ left over and we can make a group of $3$ numbers adding to $15$ with $(1, 6, 8)$ or $(2, 4, 9)$ . Another arrangement is $(3, 5, 7) (1, 6, 8) (2, 4, 9)$ . Finally, the last group has $(1, 2, 3, 7, 8, 9)$ left over. There is no way to make a group of $3$ numbers adding to $15$ with this, so the arrangements are $(1, 5, 9) (3, 4, 8) (2, 6, 7)$ and $(3, 5, 7) (1, 6, 8) (2, 4, 9)$ . So，there are $\boxed{2}$ sets that can be formed.",227176,https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_22,1,"In a sequence of positive integers, each term after the second is the product of the previous two terms. The sixth term is $4000$ . What is the first term?272$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 10$","In this solution, we will use trial and error to solve. $4000$ can be expressed as $200 \times 20$ . We divide $200$ by $20$ and get $10$ , divide $20$ by $10$ and get $2$ , and divide $10$ by $2$ to get $\boxed{5}$ . No one said that they have to be in ascending order!",527377,https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_25,1,"Fifteen integers $a_1, a_2, a_3, \dots, a_{15}$ are arranged in order on a number line. The integers are equally spaced and have the property that \[1 \le a_1 \le 10, \thickspace 13 \le a_2 \le 20, \thickspace \text{ and } \thickspace 241 \le a_{15}\le 250.\] What is the sum of digits of $a_{14}?$274$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 11 \qquad \textbf{(E)}\ 12$","We can find the possible values of the common difference by finding the numbers which satisfy the conditions. To do this, we find the minimum of the last two: $241-20=221$ , and the maximum– $250-13=237$ . There is a difference of $13$ between them, so only $17$ and $18$ work, as $17\cdot13=221$ , so $17$ satisfies $221\leq 13x\leq237$ . The number $18$ is similarly found. $19$ , however, is too much.275Now, we check with the first and last equations using the same method. We know $241-10\leq 14x\leq250-1$ . Therefore, $231\leq 14x\leq249$ . We test both values we just got, and we can realize that $18$ is too large to satisfy this inequality. On the other hand, we can now find that the difference will be $17$ , which satisfies this inequality.276The last step is to find the first term. We know that the first term can only be from $1$ to $3$ since any larger value would render the second inequality invalid. Testing these three, we find that only $a_1=3$ will satisfy all the inequalities. Therefore, $a_{14}=13\cdot17+3=224$ . The sum of the digits is therefore $\boxed{8}$",827778,https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_25,2,"Fifteen integers $a_1, a_2, a_3, \dots, a_{15}$ are arranged in order on a number line. The integers are equally spaced and have the property that \[1 \le a_1 \le 10, \thickspace 13 \le a_2 \le 20, \thickspace \text{ and } \thickspace 241 \le a_{15}\le 250.\] What is the sum of digits of $a_{14}?$278$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 11 \qquad \textbf{(E)}\ 12$","Let the common difference between consecutive $a_i$ be $d$ . 279Since $a_{15} - a_1 = 14d$ , we find from the first and last inequalities that $231 \le 14d \le 249$ . As $d$ must be an integer, this means $d = 17$ . Substituting this into all of the given inequalities so we may extract information about $a_1$ gives \[1 \le a_1 \le 10, \thickspace 13 \le a_1 + 17 \le 20, \thickspace 241 \le a_1 + 238 \le 250.\] The second inequality tells us that $1 \le a_1 \le 3$ while the last inequality tells us $3 \le a_1 \le 12$ , so we must have $a_1 = 3$ . Finally, to solve for $a_{14}$ , we simply have $a_{14} = a_1 + 13d = 3 + 13(17) = 3 + 221 = 224$ , so our answer is $2 + 2 + 4 = \boxed{8}$",828079,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_1,1,"The Math Team designed a logo shaped like a multiplication symbol, shown below on a grid of 1-inch squares. What is the area of the logo in square inches?281[asy] defaultpen(linewidth(0.5)); size(5cm); defaultpen(fontsize(14pt)); label(""$\textbf{Math}$"", (2.1,3.7)--(3.9,3.7)); label(""$\textbf{Team}$"", (2.1,3)--(3.9,3)); filldraw((1,2)--(2,1)--(3,2)--(4,1)--(5,2)--(4,3)--(5,4)--(4,5)--(3,4)--(2,5)--(1,4)--(2,3)--(1,2)--cycle, mediumgray*0.5 + lightgray*0.5);  draw((0,0)--(6,0), gray); draw((0,1)--(6,1), gray); draw((0,2)--(6,2), gray); draw((0,3)--(6,3), gray); draw((0,4)--(6,4), gray); draw((0,5)--(6,5), gray); draw((0,6)--(6,6), gray);  draw((0,0)--(0,6), gray); draw((1,0)--(1,6), gray); draw((2,0)--(2,6), gray); draw((3,0)--(3,6), gray); draw((4,0)--(4,6), gray); draw((5,0)--(5,6), gray); draw((6,0)--(6,6), gray); [/asy]282$\textbf{(A) } 10 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 14 \qquad \textbf{(E) } 15$","Draw the following four lines as shown: [asy] usepackage(""mathptmx""); defaultpen(linewidth(0.5)); size(5cm); defaultpen(fontsize(14pt)); label(""$\textbf{Math}$"", (2.1,3.7)--(3.9,3.7)); label(""$\textbf{Team}$"", (2.1,3)--(3.9,3)); filldraw((1,2)--(2,1)--(3,2)--(4,1)--(5,2)--(4,3)--(5,4)--(4,5)--(3,4)--(2,5)--(1,4)--(2,3)--(1,2)--cycle, mediumgray*0.5 + lightgray*0.5);  draw((0,0)--(6,0), gray); draw((0,1)--(6,1), gray); draw((0,2)--(6,2), gray); draw((0,3)--(6,3), gray); draw((0,4)--(6,4), gray); draw((0,5)--(6,5), gray); draw((0,6)--(6,6), gray);  draw((0,0)--(0,6), gray); draw((1,0)--(1,6), gray); draw((2,0)--(2,6), gray); draw((3,0)--(3,6), gray); draw((4,0)--(4,6), gray); draw((5,0)--(5,6), gray); draw((6,0)--(6,6), gray);  draw((3,4)--(4,3), red); draw((4,3)--(3,2), red); draw((3,2)--(2,3), red); draw((2,3)--(3,4), red); [/asy]283We see these lines split the figure into five squares with side length $\sqrt2$ . Thus, the area is $5\cdot\left(\sqrt2\right)^2=5\cdot 2 = \boxed{10}$",1028480,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_1,2,"The Math Team designed a logo shaped like a multiplication symbol, shown below on a grid of 1-inch squares. What is the area of the logo in square inches?285[asy] defaultpen(linewidth(0.5)); size(5cm); defaultpen(fontsize(14pt)); label(""$\textbf{Math}$"", (2.1,3.7)--(3.9,3.7)); label(""$\textbf{Team}$"", (2.1,3)--(3.9,3)); filldraw((1,2)--(2,1)--(3,2)--(4,1)--(5,2)--(4,3)--(5,4)--(4,5)--(3,4)--(2,5)--(1,4)--(2,3)--(1,2)--cycle, mediumgray*0.5 + lightgray*0.5);  draw((0,0)--(6,0), gray); draw((0,1)--(6,1), gray); draw((0,2)--(6,2), gray); draw((0,3)--(6,3), gray); draw((0,4)--(6,4), gray); draw((0,5)--(6,5), gray); draw((0,6)--(6,6), gray);  draw((0,0)--(0,6), gray); draw((1,0)--(1,6), gray); draw((2,0)--(2,6), gray); draw((3,0)--(3,6), gray); draw((4,0)--(4,6), gray); draw((5,0)--(5,6), gray); draw((6,0)--(6,6), gray); [/asy]286$\textbf{(A) } 10 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 14 \qquad \textbf{(E) } 15$","There are $5$ lattice points in the interior of the logo and $12$ lattice points on the boundary of the logo. Because of Pick's Theorem, the area of the logo is $5+\frac{12}{2}-1=\boxed{10}$",1028781,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_1,3,"The Math Team designed a logo shaped like a multiplication symbol, shown below on a grid of 1-inch squares. What is the area of the logo in square inches?288[asy] defaultpen(linewidth(0.5)); size(5cm); defaultpen(fontsize(14pt)); label(""$\textbf{Math}$"", (2.1,3.7)--(3.9,3.7)); label(""$\textbf{Team}$"", (2.1,3)--(3.9,3)); filldraw((1,2)--(2,1)--(3,2)--(4,1)--(5,2)--(4,3)--(5,4)--(4,5)--(3,4)--(2,5)--(1,4)--(2,3)--(1,2)--cycle, mediumgray*0.5 + lightgray*0.5);  draw((0,0)--(6,0), gray); draw((0,1)--(6,1), gray); draw((0,2)--(6,2), gray); draw((0,3)--(6,3), gray); draw((0,4)--(6,4), gray); draw((0,5)--(6,5), gray); draw((0,6)--(6,6), gray);  draw((0,0)--(0,6), gray); draw((1,0)--(1,6), gray); draw((2,0)--(2,6), gray); draw((3,0)--(3,6), gray); draw((4,0)--(4,6), gray); draw((5,0)--(5,6), gray); draw((6,0)--(6,6), gray); [/asy]289$\textbf{(A) } 10 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 14 \qquad \textbf{(E) } 15$","Notice that the area of the figure is equal to the area of the $4 \times 4$ square subtracted by the $12$ triangles that are half the area of each square, which is $1$ . The total area of the triangles not in the figure is $12 \cdot \frac{1}{2} = 6$ , so the answer is $16-6 = \boxed{10}$",1029082,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_1,4,"The Math Team designed a logo shaped like a multiplication symbol, shown below on a grid of 1-inch squares. What is the area of the logo in square inches?291[asy] defaultpen(linewidth(0.5)); size(5cm); defaultpen(fontsize(14pt)); label(""$\textbf{Math}$"", (2.1,3.7)--(3.9,3.7)); label(""$\textbf{Team}$"", (2.1,3)--(3.9,3)); filldraw((1,2)--(2,1)--(3,2)--(4,1)--(5,2)--(4,3)--(5,4)--(4,5)--(3,4)--(2,5)--(1,4)--(2,3)--(1,2)--cycle, mediumgray*0.5 + lightgray*0.5);  draw((0,0)--(6,0), gray); draw((0,1)--(6,1), gray); draw((0,2)--(6,2), gray); draw((0,3)--(6,3), gray); draw((0,4)--(6,4), gray); draw((0,5)--(6,5), gray); draw((0,6)--(6,6), gray);  draw((0,0)--(0,6), gray); draw((1,0)--(1,6), gray); draw((2,0)--(2,6), gray); draw((3,0)--(3,6), gray); draw((4,0)--(4,6), gray); draw((5,0)--(5,6), gray); draw((6,0)--(6,6), gray); [/asy]292$\textbf{(A) } 10 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 14 \qquad \textbf{(E) } 15$","Draw the following four lines as shown:293[asy] usepackage(""mathptmx""); defaultpen(linewidth(0.5)); size(5cm); defaultpen(fontsize(14pt)); label(""$\textbf{Math}$"", (2.1,3.7)--(3.9,3.7)); label(""$\textbf{Team}$"", (2.1,3)--(3.9,3)); filldraw((1,2)--(2,1)--(3,2)--(4,1)--(5,2)--(4,3)--(5,4)--(4,5)--(3,4)--(2,5)--(1,4)--(2,3)--(1,2)--cycle, mediumgray*0.5 + lightgray*0.5);  draw((0,0)--(6,0), gray); draw((0,1)--(6,1), gray); draw((0,2)--(6,2), gray); draw((0,3)--(6,3), gray); draw((0,4)--(6,4), gray); draw((0,5)--(6,5), gray); draw((0,6)--(6,6), gray);  draw((0,0)--(0,6), gray); draw((1,0)--(1,6), gray); draw((2,0)--(2,6), gray); draw((3,0)--(3,6), gray); draw((4,0)--(4,6), gray); draw((5,0)--(5,6), gray); draw((6,0)--(6,6), gray);  draw((2,4)--(4,4), red); draw((4,4)--(4,2), red); draw((4,2)--(2,2), red); draw((2,2)--(2,4), red); [/asy]294The area of the big square is $4$ , and the area of each triangle is $0.5$ . There are $12$ of these triangles, so the total area of all the triangles is $0.5\cdot12=6$ . Therefore, the area of the entire figure is $4+6=\boxed{10}$",1029583,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_1,5,"The Math Team designed a logo shaped like a multiplication symbol, shown below on a grid of 1-inch squares. What is the area of the logo in square inches?296[asy] defaultpen(linewidth(0.5)); size(5cm); defaultpen(fontsize(14pt)); label(""$\textbf{Math}$"", (2.1,3.7)--(3.9,3.7)); label(""$\textbf{Team}$"", (2.1,3)--(3.9,3)); filldraw((1,2)--(2,1)--(3,2)--(4,1)--(5,2)--(4,3)--(5,4)--(4,5)--(3,4)--(2,5)--(1,4)--(2,3)--(1,2)--cycle, mediumgray*0.5 + lightgray*0.5);  draw((0,0)--(6,0), gray); draw((0,1)--(6,1), gray); draw((0,2)--(6,2), gray); draw((0,3)--(6,3), gray); draw((0,4)--(6,4), gray); draw((0,5)--(6,5), gray); draw((0,6)--(6,6), gray);  draw((0,0)--(0,6), gray); draw((1,0)--(1,6), gray); draw((2,0)--(2,6), gray); draw((3,0)--(3,6), gray); draw((4,0)--(4,6), gray); draw((5,0)--(5,6), gray); draw((6,0)--(6,6), gray); [/asy]297$\textbf{(A) } 10 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 14 \qquad \textbf{(E) } 15$","The coordinates are $(1,2), (2,1), (3,2), (4,1), (5,2), (4,3), (5,4), (4,5), (3,4), (2,5), (1,4), (2,3)$ Use the Shoelace Theorem to get $\boxed{10}$",1029884,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_1,6,"The Math Team designed a logo shaped like a multiplication symbol, shown below on a grid of 1-inch squares. What is the area of the logo in square inches?299[asy] defaultpen(linewidth(0.5)); size(5cm); defaultpen(fontsize(14pt)); label(""$\textbf{Math}$"", (2.1,3.7)--(3.9,3.7)); label(""$\textbf{Team}$"", (2.1,3)--(3.9,3)); filldraw((1,2)--(2,1)--(3,2)--(4,1)--(5,2)--(4,3)--(5,4)--(4,5)--(3,4)--(2,5)--(1,4)--(2,3)--(1,2)--cycle, mediumgray*0.5 + lightgray*0.5);  draw((0,0)--(6,0), gray); draw((0,1)--(6,1), gray); draw((0,2)--(6,2), gray); draw((0,3)--(6,3), gray); draw((0,4)--(6,4), gray); draw((0,5)--(6,5), gray); draw((0,6)--(6,6), gray);  draw((0,0)--(0,6), gray); draw((1,0)--(1,6), gray); draw((2,0)--(2,6), gray); draw((3,0)--(3,6), gray); draw((4,0)--(4,6), gray); draw((5,0)--(5,6), gray); draw((6,0)--(6,6), gray); [/asy]300$\textbf{(A) } 10 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 14 \qquad \textbf{(E) } 15$","If the triangles are rearranged such that the gaps are filled, there would be a $4$ by $2$ rectangle, and two $1$ by $1$ squares are present. Thus, the answer is $\boxed{10}$",1030185,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_3,1,"When three positive integers $a$ $b$ , and $c$ are multiplied together, their product is $100$ . Suppose $a < b < c$ . In how many ways can the numbers be chosen?302$\textbf{(A) } 0 \qquad \textbf{(B) } 1\qquad\textbf{(C) } 2\qquad\textbf{(D) } 3\qquad\textbf{(E) } 4$","The positive divisors of $100$ are \[1,2,4,5,10,20,25,50,100.\] It is clear that $10\leq c\leq50,$ so we apply casework to $c:$303Together, the numbers $a,b,$ and $c$ can be chosen in $\boxed{4}$ ways.",430486,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_3,2,"When three positive integers $a$ $b$ , and $c$ are multiplied together, their product is $100$ . Suppose $a < b < c$ . In how many ways can the numbers be chosen?305$\textbf{(A) } 0 \qquad \textbf{(B) } 1\qquad\textbf{(C) } 2\qquad\textbf{(D) } 3\qquad\textbf{(E) } 4$","The positive divisors of $100$ are \[1,2,4,5,10,20,25,50,100.\] We apply casework to $a$306If $a=1$ , then there are $3$ cases:307If $a=2$ , then there is only $1$ case:308In total, there are $3+1=\boxed{4}$ ways to choose distinct positive integer values of $a,b,c$",430987,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_5,1,"Anna and Bella are celebrating their birthdays together. Five years ago, when Bella turned $6$ years old, she received a newborn kitten as a birthday present. Today the sum of the ages of the two children and the kitten is $30$ years. How many years older than Bella is Anna?310$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } ~5$","Five years ago, Bella was $6$ years old, and the kitten was $0$ years old.311Today, Bella is $11$ years old, and the kitten is $5$ years old. It follows that Anna is $30-11-5=14$ years old.312Therefore, Anna is $14-11=\boxed{3}$ years older than Bella.",331388,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_6,1,"Three positive integers are equally spaced on a number line. The middle number is $15,$ and the largest number is $4$ times the smallest number. What is the smallest of these three numbers?314$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$","Let the smallest number be $x.$ It follows that the largest number is $4x.$315Since $x,15,$ and $4x$ are equally spaced on a number line, we have \begin{align*} 4x-15 &= 15-x \\ 5x &= 30 \\ x &= \boxed{6} ~MRENTHUSIASM",631689,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_6,2,"Three positive integers are equally spaced on a number line. The middle number is $15,$ and the largest number is $4$ times the smallest number. What is the smallest of these three numbers?317$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$","Let the common difference of the arithmetic sequence be $d$ . Consequently, the smallest number is $15-d$ and the largest number is $15+d$ . As the largest number is $4$ times the smallest number, $15+d=60-4d\implies d=9$ . Finally, we find that the smallest number is $15-9=\boxed{6}$",631890,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_6,3,"Three positive integers are equally spaced on a number line. The middle number is $15,$ and the largest number is $4$ times the smallest number. What is the smallest of these three numbers?319$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$","Let the smallest number be $x$ . Since the integers are equally spaced, and there are three of them, the middle number ( $15$ ) is the arithmetic mean of the other two numbers ( $x$ and $4x$ ). Thus, we set up the equation $(4x + x)/3 = 15$ , and, solving for $x$ , get $x = 6$ . Since $6$ is the smallest number out of the list $6, 15, 24$ $24$ because it equals $4x$ ), the answer is $\boxed{6}$",632091,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_6,4,"Three positive integers are equally spaced on a number line. The middle number is $15,$ and the largest number is $4$ times the smallest number. What is the smallest of these three numbers?321$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$","Let the smallest number be $x$ . Because $x$ and $4x$ are equally spaced from $15$ $15$ must be the average. By adding $x$ and $4x$ and dividing by $2$ , we get that the mean is also $2.5x$ . We get that $2.5x=15$ , and solving gets $x=\boxed{6}$",632292,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_7,1,"When the World Wide Web first became popular in the $1990$ s, download speeds reached a maximum of about $56$ kilobits per second. Approximately how many minutes would the download of a $4.2$ -megabyte song have taken at that speed? (Note that there are $8000$ kilobits in a megabyte.)323$\textbf{(A) } 0.6 \qquad \textbf{(B) } 10 \qquad \textbf{(C) } 1800 \qquad \textbf{(D) } 7200 \qquad \textbf{(E) } 36000$","Notice that the number of kilobits in this song is $4.2 \cdot 8000 = 8 \cdot 7 \cdot 6 \cdot 100.$324We must divide this by $56$ in order to find out how many seconds this song would take to download: $\frac{\cancel{8}\cdot\cancel{7}\cdot6\cdot100}{\cancel{56}} = 600.$325Finally, we divide  this number by $60$ because this is the number of seconds to get the answer $\frac{600}{60}=\boxed{10}.$",1032693,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_7,2,"When the World Wide Web first became popular in the $1990$ s, download speeds reached a maximum of about $56$ kilobits per second. Approximately how many minutes would the download of a $4.2$ -megabyte song have taken at that speed? (Note that there are $8000$ kilobits in a megabyte.)327$\textbf{(A) } 0.6 \qquad \textbf{(B) } 10 \qquad \textbf{(C) } 1800 \qquad \textbf{(D) } 7200 \qquad \textbf{(E) } 36000$","We seek a value of $x$ that makes the following equation true, since every other quantity equals $1$328\[\frac{x\ \text{min}}{4.2\ \text{mb}} \cdot \frac{56\ \text{kb}}{1\ \text{sec}} \cdot \frac{1\ \text{mb}}{8000\ \text{kb}} \cdot \frac{60\ \text{sec}}{1\ \text{min}} = 1.\] Solving yields $x=\boxed{10}$",1032994,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_9,1,"A cup of boiling water ( $212^{\circ}\text{F}$ ) is placed to cool in a room whose temperature remains constant at $68^{\circ}\text{F}$ . Suppose the difference between the water temperature and the room temperature is halved every $5$ minutes. What is the water temperature, in degrees Fahrenheit, after $15$ minutes?330$\textbf{(A) } 77 \qquad \textbf{(B) } 86 \qquad \textbf{(C) } 92 \qquad \textbf{(D) } 98 \qquad \textbf{(E) } 104$","Initially, the difference between the water temperature and the room temperature is $212-68=144$ degrees Fahrenheit.331After $5$ minutes, the difference between the temperatures is $144\div2=72$ degrees Fahrenheit.332After $10$ minutes, the difference between the temperatures is $72\div2=36$ degrees Fahrenheit.333After $15$ minutes, the difference between the temperatures is $36\div2=18$ degrees Fahrenheit. At this point, the water temperature is $68+18=\boxed{86}$ degrees Fahrenheit.",8633495,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_11,1,"Henry the donkey has a very long piece of pasta. He takes a number of bites of pasta, each time eating $3$ inches of pasta from the middle of one piece. In the end, he has $10$ pieces of pasta whose total length is $17$ inches. How long, in inches, was the piece of pasta he started with?335$\textbf{(A) } 34\qquad\textbf{(B) } 38\qquad\textbf{(C) } 41\qquad\textbf{(D) } 44\qquad\textbf{(E) } 47$","If there are $10$ pieces of pasta, Henry took $10-1=9$ bites. Each of these $9$ bites took $3$ inches of pasta out, and thus his bites in total took away $9\cdot 3 = 27$ inches of pasta. Thus, the original piece of pasta was $27+17=\boxed{44}$ inches long.",4433696,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_13,1,"How many positive integers can fill the blank in the sentence below?337“One positive integer is _____ more than twice another, and the sum of the two numbers is $28$ .”338$\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 10$","Let $m$ and $n$ be positive integers such that $m>n$ and $m+n=28.$ It follows that $m=2n+d$ for some positive integer $d.$ We wish to find the number of possible values for $d.$339By substitution, we have $(2n+d)+n=28,$ from which $d=28-3n.$ Note that $n=1,2,3,\ldots,9$ each generate a positive integer for $d,$ so there are $\boxed{9}$ possible values for $d.$",934097,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_14,1,"In how many ways can the letters in $\textbf{BEEKEEPER}$ be rearranged so that two or more $\textbf{E}$ s do not appear together?341$\textbf{(A) } 1 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 24 \qquad \textbf{(E) } 120$","All valid arrangements of the letters must be of the form \[\textbf{E\underline{\hspace{3mm}}E\underline{\hspace{3mm}}E\underline{\hspace{3mm}}E\underline{\hspace{3mm}}E}.\] The problem is equivalent to counting the arrangements of $\textbf{B},\textbf{K},\textbf{P},$ and $\textbf{R}$ into the four blanks, in which there are $4!=\boxed{24}$ ways.",2434298,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_15,1,"Laszlo went online to shop for black pepper and found thirty different black pepper options varying in weight and price, shown in the scatter plot below. In ounces, what is the weight of the pepper that offers the lowest price per ounce?343[asy] //diagram by pog size(5.5cm); usepackage(""mathptmx""); defaultpen(mediumgray*0.5+gray*0.5+linewidth(0.63)); add(grid(6,6)); label(scale(0.7)*""$1$"", (1,-0.3), black); label(scale(0.7)*""$2$"", (2,-0.3), black); label(scale(0.7)*""$3$"", (3,-0.3), black); label(scale(0.7)*""$4$"", (4,-0.3), black); label(scale(0.7)*""$5$"", (5,-0.3), black); label(scale(0.7)*""$1$"", (-0.3,1), black); label(scale(0.7)*""$2$"", (-0.3,2), black); label(scale(0.7)*""$3$"", (-0.3,3), black); label(scale(0.7)*""$4$"", (-0.3,4), black); label(scale(0.7)*""$5$"", (-0.3,5), black); label(scale(0.8)*rotate(90)*""Price (dollars)"", (-1,3.2), black); label(scale(0.8)*""Weight (ounces)"", (3.2,-1), black); dot((1,1.2),black); dot((1,1.7),black); dot((1,2),black); dot((1,2.8),black);  dot((1.5,2.1),black); dot((1.5,3),black); dot((1.5,3.3),black); dot((1.5,3.75),black);  dot((2,2),black); dot((2,2.9),black); dot((2,3),black); dot((2,4),black); dot((2,4.35),black); dot((2,4.8),black);  dot((2.5,2.7),black); dot((2.5,3.7),black); dot((2.5,4.2),black); dot((2.5,4.4),black);  dot((3,2.5),black); dot((3,3.4),black); dot((3,4.2),black);  dot((3.5,3.8),black); dot((3.5,4.5),black); dot((3.5,4.8),black);  dot((4,3.9),black); dot((4,5.1),black);  dot((4.5,4.75),black); dot((4.5,5),black);  dot((5,4.5),black); dot((5,5),black); [/asy]344$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }5$","[asy] //diagram by pog size(5.5cm); usepackage(""mathptmx""); defaultpen(mediumgray*0.5+gray*0.5+linewidth(0.63)); add(grid(6,6)); label(scale(0.7)*""$1$"", (1,-0.3), black); label(scale(0.7)*""$2$"", (2,-0.3), black); label(scale(0.7)*""$3$"", (3,-0.3), black); label(scale(0.7)*""$4$"", (4,-0.3), black); label(scale(0.7)*""$5$"", (5,-0.3), black); label(scale(0.7)*""$1$"", (-0.3,1), black); label(scale(0.7)*""$2$"", (-0.3,2), black); label(scale(0.7)*""$3$"", (-0.3,3), black); label(scale(0.7)*""$4$"", (-0.3,4), black); label(scale(0.7)*""$5$"", (-0.3,5), black); label(scale(0.8)*rotate(90)*""Price (dollars)"", (-1,3.2), black); label(scale(0.8)*""Weight (ounces)"", (3.2,-1), black); draw((0,0)--(6,5),red);  dot((1,1.2),black); dot((1,1.7),black); dot((1,2),black); dot((1,2.8),black);  dot((1.5,2.1),black); dot((1.5,3),black); dot((1.5,3.3),black); dot((1.5,3.75),black);  dot((2,2),black); dot((2,2.9),black); dot((2,3),black); dot((2,4),black); dot((2,4.35),black); dot((2,4.8),black);  dot((2.5,2.7),black); dot((2.5,3.7),black); dot((2.5,4.2),black); dot((2.5,4.4),black);  dot((3,2.5),blue); dot((3,3.4),black); dot((3,4.2),black);  dot((3.5,3.8),black); dot((3.5,4.5),black); dot((3.5,4.8),black);  dot((4,3.9),black); dot((4,5.1),black);  dot((4.5,4.75),black); dot((4.5,5),black);  dot((5,4.5),black); dot((5,5),black); [/asy]345We are looking for a black point, that when connected to the origin, yields the lowest slope. The slope represents the price per ounce. We can visually find that the point with the lowest slope is the blue point. Furthermore, it is the only one with a price per ounce significantly less than $1$ . Finally, we see that the blue point is in the category with a weight of $\boxed{3}$ ounces.",334699,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_15,2,"Laszlo went online to shop for black pepper and found thirty different black pepper options varying in weight and price, shown in the scatter plot below. In ounces, what is the weight of the pepper that offers the lowest price per ounce?347[asy] //diagram by pog size(5.5cm); usepackage(""mathptmx""); defaultpen(mediumgray*0.5+gray*0.5+linewidth(0.63)); add(grid(6,6)); label(scale(0.7)*""$1$"", (1,-0.3), black); label(scale(0.7)*""$2$"", (2,-0.3), black); label(scale(0.7)*""$3$"", (3,-0.3), black); label(scale(0.7)*""$4$"", (4,-0.3), black); label(scale(0.7)*""$5$"", (5,-0.3), black); label(scale(0.7)*""$1$"", (-0.3,1), black); label(scale(0.7)*""$2$"", (-0.3,2), black); label(scale(0.7)*""$3$"", (-0.3,3), black); label(scale(0.7)*""$4$"", (-0.3,4), black); label(scale(0.7)*""$5$"", (-0.3,5), black); label(scale(0.8)*rotate(90)*""Price (dollars)"", (-1,3.2), black); label(scale(0.8)*""Weight (ounces)"", (3.2,-1), black); dot((1,1.2),black); dot((1,1.7),black); dot((1,2),black); dot((1,2.8),black);  dot((1.5,2.1),black); dot((1.5,3),black); dot((1.5,3.3),black); dot((1.5,3.75),black);  dot((2,2),black); dot((2,2.9),black); dot((2,3),black); dot((2,4),black); dot((2,4.35),black); dot((2,4.8),black);  dot((2.5,2.7),black); dot((2.5,3.7),black); dot((2.5,4.2),black); dot((2.5,4.4),black);  dot((3,2.5),black); dot((3,3.4),black); dot((3,4.2),black);  dot((3.5,3.8),black); dot((3.5,4.5),black); dot((3.5,4.8),black);  dot((4,3.9),black); dot((4,5.1),black);  dot((4.5,4.75),black); dot((4.5,5),black);  dot((5,4.5),black); dot((5,5),black); [/asy]348$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }5$","By the answer choices, we can disregard the points that do not have integer weights. As a result, we obtain the following diagram:349[asy] //diagram by pog size(5.5cm); usepackage(""mathptmx""); defaultpen(mediumgray*0.5+gray*0.5+linewidth(0.63)); add(grid(6,6)); label(scale(0.7)*""$1$"", (1,-0.3), black); label(scale(0.7)*""$2$"", (2,-0.3), black); label(scale(0.7)*""$3$"", (3,-0.3), black); label(scale(0.7)*""$4$"", (4,-0.3), black); label(scale(0.7)*""$5$"", (5,-0.3), black); label(scale(0.7)*""$1$"", (-0.3,1), black); label(scale(0.7)*""$2$"", (-0.3,2), black); label(scale(0.7)*""$3$"", (-0.3,3), black); label(scale(0.7)*""$4$"", (-0.3,4), black); label(scale(0.7)*""$5$"", (-0.3,5), black); label(scale(0.8)*rotate(90)*""Price (dollars)"", (-1,3.2), black); label(scale(0.8)*""Weight (ounces)"", (3.2,-1), black); dot((1,1.2),black); dot((1,1.7),black); dot((1,2),black); dot((1,2.8),black);  dot((2,2),black); dot((2,2.9),black); dot((2,3),black); dot((2,4),black); dot((2,4.35),black); dot((2,4.8),black);  dot((3,2.5),blue); dot((3,3.4),black); dot((3,4.2),black);  dot((4,3.9),black); dot((4,5.1),black);  dot((5,4.5),black); dot((5,5),black); [/asy]350We then proceed in the same way that we had done in Solution 1. Following the steps, we figure out the blue dot that yields the lowest slope, along with passing the origin. We then can look at the x-axis(in this situation, the weight) and figure out it has $\boxed{3}$ ounces.",3351100,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_15,3,"Laszlo went online to shop for black pepper and found thirty different black pepper options varying in weight and price, shown in the scatter plot below. In ounces, what is the weight of the pepper that offers the lowest price per ounce?352[asy] //diagram by pog size(5.5cm); usepackage(""mathptmx""); defaultpen(mediumgray*0.5+gray*0.5+linewidth(0.63)); add(grid(6,6)); label(scale(0.7)*""$1$"", (1,-0.3), black); label(scale(0.7)*""$2$"", (2,-0.3), black); label(scale(0.7)*""$3$"", (3,-0.3), black); label(scale(0.7)*""$4$"", (4,-0.3), black); label(scale(0.7)*""$5$"", (5,-0.3), black); label(scale(0.7)*""$1$"", (-0.3,1), black); label(scale(0.7)*""$2$"", (-0.3,2), black); label(scale(0.7)*""$3$"", (-0.3,3), black); label(scale(0.7)*""$4$"", (-0.3,4), black); label(scale(0.7)*""$5$"", (-0.3,5), black); label(scale(0.8)*rotate(90)*""Price (dollars)"", (-1,3.2), black); label(scale(0.8)*""Weight (ounces)"", (3.2,-1), black); dot((1,1.2),black); dot((1,1.7),black); dot((1,2),black); dot((1,2.8),black);  dot((1.5,2.1),black); dot((1.5,3),black); dot((1.5,3.3),black); dot((1.5,3.75),black);  dot((2,2),black); dot((2,2.9),black); dot((2,3),black); dot((2,4),black); dot((2,4.35),black); dot((2,4.8),black);  dot((2.5,2.7),black); dot((2.5,3.7),black); dot((2.5,4.2),black); dot((2.5,4.4),black);  dot((3,2.5),black); dot((3,3.4),black); dot((3,4.2),black);  dot((3.5,3.8),black); dot((3.5,4.5),black); dot((3.5,4.8),black);  dot((4,3.9),black); dot((4,5.1),black);  dot((4.5,4.75),black); dot((4.5,5),black);  dot((5,4.5),black); dot((5,5),black); [/asy]353$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }5$","We can find the lowest point in each line ( $1$ $2$ $3$ $4$ , or $5$ ) and find the price per pound. (Note that we don't need to find the points higher than the points below since we are finding the lowest price per pound.)354[asy] //diagram by pog size(5.5cm); usepackage(""mathptmx""); defaultpen(mediumgray*0.5+gray*0.5+linewidth(0.63)); add(grid(6,6)); label(scale(0.7)*""$1$"", (1,-0.3), black); label(scale(0.7)*""$2$"", (2,-0.3), black); label(scale(0.7)*""$3$"", (3,-0.3), black); label(scale(0.7)*""$4$"", (4,-0.3), black); label(scale(0.7)*""$5$"", (5,-0.3), black); label(scale(0.7)*""$1$"", (-0.3,1), black); label(scale(0.7)*""$2$"", (-0.3,2), black); label(scale(0.7)*""$3$"", (-0.3,3), black); label(scale(0.7)*""$4$"", (-0.3,4), black); label(scale(0.7)*""$5$"", (-0.3,5), black); label(scale(0.8)*rotate(90)*""Price (dollars)"", (-1,3.2), black); label(scale(0.8)*""Weight (ounces)"", (3.2,-1), black); dot((1,1.2),red); dot((1,1.7),black); dot((1,2),black); dot((1,2.8),black);  dot((2,2),green); dot((2,2.9),black); dot((2,3),black); dot((2,4),black); dot((2,4.35),black); dot((2,4.8),black);  dot((3,2.5),blue); dot((3,3.4),black); dot((3,4.2),black);  dot((4,3.9),orange); dot((4,5.1),black);  dot((5,4.5),purple); dot((5,5),black); [/asy]355The red dot has a price per pound of something that is larger than $1$ . The green dot has a price per pound of $1$ . The blue dot has a price per pound of something like $\frac{2.5}{3}$ . The orange dot has a price per pound that is less than $1$ , but is very close to it. The purple dot has a price per pound of something like $\frac{4.5}{5}$ . We see that choices $\textbf{(A)}$ $\textbf{(B)}$ ,and $\textbf{(D)}$ are eliminated. Also, $\frac{4.5}{5} > \frac{2.5}{3}$ thus the answer is $\boxed{3}$",3356101,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_16,1,"Four numbers are written in a row. The average of the first two is $21,$ the average of the middle two is $26,$ and the average of the last two is $30.$ What is the average of the first and last of the numbers?357$\textbf{(A) } 24 \qquad \textbf{(B) } 25 \qquad \textbf{(C) } 26 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 28$","Note that the sum of the first two numbers is $21\cdot2=42,$ the sum of the middle two numbers is $26\cdot2=52,$ and the sum of the last two numbers is $30\cdot2=60.$358It follows that the sum of the four numbers is $42+60=102,$ so the sum of the first and last numbers is $102-52=50.$ Therefore, the average of the first and last numbers is $50\div2=\boxed{25}.$",25359102,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_16,2,"Four numbers are written in a row. The average of the first two is $21,$ the average of the middle two is $26,$ and the average of the last two is $30.$ What is the average of the first and last of the numbers?360$\textbf{(A) } 24 \qquad \textbf{(B) } 25 \qquad \textbf{(C) } 26 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 28$","Let $a,b,c,$ and $d$ be the four numbers in that order. We are given that \begin{align*} \frac{a+b}{2} &= 21, &(1) \\ \frac{b+c}{2} &= 26, &(2) \\ \frac{c+d}{2} &= 30, &(3) \end{align*} and we wish to find $\frac{a+d}{2}.$361We add $(1)$ and $(3),$ then subtract $(2)$ from the result: \[\frac{a+d}{2}=21+30-26=\boxed{25}.\] ~MRENTHUSIASM",25362103,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_16,3,"Four numbers are written in a row. The average of the first two is $21,$ the average of the middle two is $26,$ and the average of the last two is $30.$ What is the average of the first and last of the numbers?363$\textbf{(A) } 24 \qquad \textbf{(B) } 25 \qquad \textbf{(C) } 26 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 28$","We can just assume some of the numbers. For example, let the first two numbers both be $21.$ It follows that the third number is $31,$ and the fourth number is $29.$ Therefore, the average of the first and last numbers is $\dfrac{21+29}2=\dfrac{50}2=\boxed{25}.$",25364104,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_17,1,"If $n$ is an even positive integer, the $\emph{double factorial}$ notation $n!!$ represents the product of all the even integers from $2$ to $n$ . For example, $8!! = 2 \cdot 4 \cdot 6 \cdot 8$ . What is the units digit of the following sum? \[2!! + 4!! + 6!! + \cdots + 2018!! + 2020!! + 2022!!\]365$\textbf{(A) } 0\qquad\textbf{(B) } 2\qquad\textbf{(C) } 4\qquad\textbf{(D) } 6\qquad\textbf{(E) } 8$","Notice that once $n>8,$ the units digit of $n!!$ will be $0$ because there will be a factor of $10.$ Thus, we only need to calculate the units digit of \[2!!+4!!+6!!+8!! = 2+8+48+48\cdot8.\] We only care about units digits, so we have $2+8+8+8\cdot8,$ which has the same units digit as $2+8+8+4.$ The answer is $\boxed{2}.$",2366105,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_18,1,"The midpoints of the four sides of a rectangle are $(-3,0), (2,0), (5,4),$ and $(0,4).$ What is the367area of the rectangle?368$\textbf{(A) } 20 \qquad \textbf{(B) } 25 \qquad \textbf{(C) } 40 \qquad \textbf{(D) } 50 \qquad \textbf{(E) } 80$","The midpoints of the four sides of every rectangle are the vertices of a rhombus whose area is half the area of the rectangle: Note that the diagonals of the rhombus have the same lengths as the sides of the rectangle.369Let $A=(-3,0), B=(2,0), C=(5,4),$ and $D=(0,4).$ Note that $A,B,C,$ and $D$ are the vertices of a rhombus whose diagonals have lengths $AC=4\sqrt{5}$ and $BD=2\sqrt{5}.$ It follows that the dimensions of the rectangle are $4\sqrt{5}$ and $2\sqrt{5},$ so the area of the rectangle is $4\sqrt{5}\cdot2\sqrt{5}=\boxed{40}.$",40370106,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_18,2,"The midpoints of the four sides of a rectangle are $(-3,0), (2,0), (5,4),$ and $(0,4).$ What is the371area of the rectangle?372$\textbf{(A) } 20 \qquad \textbf{(B) } 25 \qquad \textbf{(C) } 40 \qquad \textbf{(D) } 50 \qquad \textbf{(E) } 80$","If a rectangle has area $K,$ then the area of the quadrilateral formed by its midpoints is $\frac{K}{2}.$373Define points $A,B,C,$ and $D$ as Solution 1 does. Since $A,B,C,$ and $D$ are the midpoints of the rectangle, the rectangle's area is $2[ABCD].$ Now, note that $ABCD$ is a parallelogram since $AB=CD$ and $\overline{AB}\parallel\overline{CD}.$ As the parallelogram's height from $D$ to $\overline{AB}$ is $4$ and $AB=5,$ its area is $4\cdot5=20.$ Therefore, the area of the rectangle is $20\cdot2=\boxed{40}.$ ~Fruitz",40374107,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_19,1,"Mr. Ramos gave a test to his class of $20$ students. The dot plot below shows the distribution of test scores. [asy] //diagram by pog . give me 1,000,000,000 dollars for this diagram size(5cm); defaultpen(0.7); dot((0.5,1)); dot((0.5,1.5)); dot((1.5,1)); dot((1.5,1.5)); dot((2.5,1)); dot((2.5,1.5)); dot((2.5,2)); dot((2.5,2.5)); dot((3.5,1)); dot((3.5,1.5)); dot((3.5,2)); dot((3.5,2.5)); dot((3.5,3)); dot((4.5,1)); dot((4.5,1.5)); dot((5.5,1)); dot((5.5,1.5)); dot((5.5,2)); dot((6.5,1)); dot((7.5,1)); draw((0,0.5)--(8,0.5),linewidth(0.7)); defaultpen(fontsize(10.5pt)); label(""$65$"", (0.5,-0.1)); label(""$70$"", (1.5,-0.1)); label(""$75$"", (2.5,-0.1)); label(""$80$"", (3.5,-0.1)); label(""$85$"", (4.5,-0.1)); label(""$90$"", (5.5,-0.1)); label(""$95$"", (6.5,-0.1)); label(""$100$"", (7.5,-0.1)); [/asy]375Later Mr. Ramos discovered that there was a scoring error on one of the questions. He regraded the tests, awarding some of the students $5$ extra points, which increased the median test score to $85$ . What is the minimum number of students who received extra points?376(Note that the median test score equals the average of the $2$ scores in the middle if the $20$ test scores are arranged in increasing order.)377$\textbf{(A)} ~2\qquad\textbf{(B)} ~3\qquad\textbf{(C)} ~4\qquad\textbf{(D)} ~5\qquad\textbf{(E)} ~6\qquad$","We set up our cases as solution 1 showed, realizing that only the second case is possible.378We notice that $13$ students have scores under $85$ currently and only $5$ have scores over $85$ . We find the median of these two numbers, getting:379\[13-5=8\] \[\frac{8}{2}=4\] \[13-4=9\]380Thus, we realize that $4$ students must have their score increased by $5$381So, the correct answer is $\boxed{4}$",4382108,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_20,1,"The grid below is to be filled with integers in such a way that the sum of the numbers in each row and the sum of the numbers in each column are the same. Four numbers are missing. The number $x$ in the lower left corner is larger than the other three missing numbers. What is the smallest possible value of $x$ [asy] unitsize(0.5cm); draw((3,3)--(-3,3)); draw((3,1)--(-3,1)); draw((3,-3)--(-3,-3)); draw((3,-1)--(-3,-1)); draw((3,3)--(3,-3)); draw((1,3)--(1,-3)); draw((-3,3)--(-3,-3)); draw((-1,3)--(-1,-3)); label((-2,2),""$-2$""); label((0,2),""$9$""); label((2,2),""$5$""); label((2,0),""$-1$""); label((2,-2),""$8$""); label((-2,-2),""$x$""); [/asy] $\textbf{(A) } -1 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 8 \qquad \textbf{(E) } 9 \qquad$","The sum of the numbers in each row is $12$ . Consider the second row. In order for the sum of the numbers in this row to equal $12$ , the two shaded numbers must add up to $13$ [asy] unitsize(0.5cm); fill((-3,1)--(1,1)--(1,-1)--(-3,-1)--cycle,mediumgray); draw((3,3)--(-3,3)); draw((3,1)--(-3,1)); draw((3,-3)--(-3,-3)); draw((3,-1)--(-3,-1)); draw((3,3)--(3,-3)); draw((1,3)--(1,-3)); draw((-3,3)--(-3,-3)); draw((-1,3)--(-1,-3)); label((-2,2),""$-2$""); label((0,2),""$9$""); label((2,2),""$5$""); label((2,0),""$-1$""); label((2,-2),""$8$""); label((-2,-2),""$x$""); [/asy] If two numbers add up to $13$ , one of them must be at least $7$ : If both shaded numbers are no more than $6$ , their sum can be at most $12$ . Therefore, for $x$ to be larger than the three missing numbers, $x$ must be at least $8$ . We can construct a working scenario where $x=8$ [asy] unitsize(0.5cm); draw((3,3)--(-3,3)); draw((3,1)--(-3,1)); draw((3,-3)--(-3,-3)); draw((3,-1)--(-3,-1)); draw((3,3)--(3,-3)); draw((1,3)--(1,-3)); draw((-3,3)--(-3,-3)); draw((-1,3)--(-1,-3)); label((-2,2),""$-2$""); label((0,2),""$9$""); label((2,2),""$5$""); label((2,0),""$-1$""); label((2,-2),""$8$""); label((-2,-2),""$8$""); label((0,-2),""$-4$""); label((-2,0),""$6$""); label((0,0),""$7$""); [/asy] So, our answer is $\boxed{8}$",8383109,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_20,2,"The grid below is to be filled with integers in such a way that the sum of the numbers in each row and the sum of the numbers in each column are the same. Four numbers are missing. The number $x$ in the lower left corner is larger than the other three missing numbers. What is the smallest possible value of $x$ [asy] unitsize(0.5cm); draw((3,3)--(-3,3)); draw((3,1)--(-3,1)); draw((3,-3)--(-3,-3)); draw((3,-1)--(-3,-1)); draw((3,3)--(3,-3)); draw((1,3)--(1,-3)); draw((-3,3)--(-3,-3)); draw((-1,3)--(-1,-3)); label((-2,2),""$-2$""); label((0,2),""$9$""); label((2,2),""$5$""); label((2,0),""$-1$""); label((2,-2),""$8$""); label((-2,-2),""$x$""); [/asy] $\textbf{(A) } -1 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 8 \qquad \textbf{(E) } 9 \qquad$","The sum of the numbers in each row is $-2+9+5=12,$ and the sum of the numbers in each column is $5+(-1)+8=12.$384Let $y$ be the number in the lower middle. It follows that $x+y+8=12,$ or $x+y=4.$385We express the other two missing numbers in terms of $x$ and $y,$ as shown below: [asy] unitsize(0.5cm); draw((3,3)--(-3,3)); draw((3,1)--(-3,1)); draw((3,-3)--(-3,-3)); draw((3,-1)--(-3,-1)); draw((3,3)--(3,-3)); draw((1,3)--(1,-3)); draw((-3,3)--(-3,-3)); draw((-1,3)--(-1,-3)); label((-2,2),""$-2$""); label((0,2),""$9$""); label((2,2),""$5$""); label((2,0),""$-1$""); label((2,-2),""$8$""); label((-2,-2),""$x$""); label((0,-2),""$y$"",red+fontsize(11)); label((-2,0),""$y{+}10$"",red+fontsize(11)); label((0,0),""$x{-}1$"",red+fontsize(11));  [/asy] We have $x>x-1, x>y+10,$ and $x>y.$ Note that the first inequality is true for all values of $x.$ We only need to solve the second inequality so that the third inequality is true for all values of $x.$ By substitution, we get $x>(4-x)+10,$ from which $x>7.$386Therefore, the smallest possible value of $x$ is $\boxed{8}.$",8387110,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_20,3,"The grid below is to be filled with integers in such a way that the sum of the numbers in each row and the sum of the numbers in each column are the same. Four numbers are missing. The number $x$ in the lower left corner is larger than the other three missing numbers. What is the smallest possible value of $x$ [asy] unitsize(0.5cm); draw((3,3)--(-3,3)); draw((3,1)--(-3,1)); draw((3,-3)--(-3,-3)); draw((3,-1)--(-3,-1)); draw((3,3)--(3,-3)); draw((1,3)--(1,-3)); draw((-3,3)--(-3,-3)); draw((-1,3)--(-1,-3)); label((-2,2),""$-2$""); label((0,2),""$9$""); label((2,2),""$5$""); label((2,0),""$-1$""); label((2,-2),""$8$""); label((-2,-2),""$x$""); [/asy] $\textbf{(A) } -1 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 8 \qquad \textbf{(E) } 9 \qquad$","This is based on the Solution 2 above and it is perhaps a little simpler than that.388Let $y$ be the number in the lower middle. Applying summation to first two columns yields the following.389[asy] unitsize(0.5cm); draw((3,3)--(-3,3)); draw((3,1)--(-3,1)); draw((3,-3)--(-3,-3)); draw((3,-1)--(-3,-1)); draw((3,3)--(3,-3)); draw((1,3)--(1,-3)); draw((-3,3)--(-3,-3)); draw((-1,3)--(-1,-3)); label((-2,2),""$-2$""); label((0,2),""$9$""); label((2,2),""$5$""); label((2,0),""$-1$""); label((2,-2),""$8$""); label((-2,-2),""$x$""); label((0,-2),""$y$"",red+fontsize(11)); label((-2,0),""$14{-}x$"",red+fontsize(11)); label((0,0),""$3{-}y$"",red+fontsize(11));  [/asy]390Since $x$ is greater than the other three, we have $x>14-x,$ or $x>7.$391Therefore, the smallest possible value of $x$ is $\boxed{8}.$",8392111,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_20,4,"The grid below is to be filled with integers in such a way that the sum of the numbers in each row and the sum of the numbers in each column are the same. Four numbers are missing. The number $x$ in the lower left corner is larger than the other three missing numbers. What is the smallest possible value of $x$ [asy] unitsize(0.5cm); draw((3,3)--(-3,3)); draw((3,1)--(-3,1)); draw((3,-3)--(-3,-3)); draw((3,-1)--(-3,-1)); draw((3,3)--(3,-3)); draw((1,3)--(1,-3)); draw((-3,3)--(-3,-3)); draw((-1,3)--(-1,-3)); label((-2,2),""$-2$""); label((0,2),""$9$""); label((2,2),""$5$""); label((2,0),""$-1$""); label((2,-2),""$8$""); label((-2,-2),""$x$""); [/asy] $\textbf{(A) } -1 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 8 \qquad \textbf{(E) } 9 \qquad$","Note that the sum of the rows and columns must be $8+5-1=12$ . We proceed to test the answer choices.393Testing $\textbf{(A)}$ , when $x = -1$ , the number above $x$ must be $15$ , which contradicts the precondition that the numbers surrounding $x$ is less than $x$394Testing $\textbf{(B)}$ , the number above $x$ is $9$ , which does not work.395Testing $\textbf{(C)}$ , the number above $x$ is $8$ , which does not work.396Testing $\textbf{(D)}$ , the number above $x$ is $6$ , which does work. Hence, the answer is $\boxed{8}$",8397112,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_20,5,"The grid below is to be filled with integers in such a way that the sum of the numbers in each row and the sum of the numbers in each column are the same. Four numbers are missing. The number $x$ in the lower left corner is larger than the other three missing numbers. What is the smallest possible value of $x$ [asy] unitsize(0.5cm); draw((3,3)--(-3,3)); draw((3,1)--(-3,1)); draw((3,-3)--(-3,-3)); draw((3,-1)--(-3,-1)); draw((3,3)--(3,-3)); draw((1,3)--(1,-3)); draw((-3,3)--(-3,-3)); draw((-1,3)--(-1,-3)); label((-2,2),""$-2$""); label((0,2),""$9$""); label((2,2),""$5$""); label((2,0),""$-1$""); label((2,-2),""$8$""); label((-2,-2),""$x$""); [/asy] $\textbf{(A) } -1 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 8 \qquad \textbf{(E) } 9 \qquad$","The sum of the numbers in each column and row should be $5+(-1)+8=12$ . If we look at the $1^{\text{st}}$ column, the gray squares (shown below) sum to $12-(-2)=14$398[asy] draw((3,3)--(-3,3)); draw((3,1)--(-3,1)); draw((3,-3)--(-3,-3)); draw((3,-1)--(-3,-1)); draw((3,3)--(3,-3)); draw((1,3)--(1,-3)); draw((-3,3)--(-3,-3)); draw((-1,3)--(-1,-3)); label((-2,2),""$-2$""); label((0,2),""$9$""); label((2,2),""$5$""); label((2,0),""$-1$""); label((2,-2),""$8$""); label((-2,-2),""$x$""); filldraw((-3,-3)--(-1,-3)--(-1,-1)--(-3,-1)--cycle, lightgray, black+linewidth(1)); filldraw((-1,-1)--(-3,-1)--(-3,1)--(-1,1)--cycle, lightgray, black+linewidth(1)); label(scale(1)*""All credits for original unedited asymptote for the problem go to whoever made the asymptote in the 'Problem' section."", (-0,-5), S); [/asy]399If square $x$ has to be greater than or equal to the three blank squares, then the least $x$ can be is half the sum of the value of the gray squares, which is $14\div2=7$ . But square $x$ has to be greater than and not greater than or equal to the three blank squares, so the least $x$ can be is $7+1=8$ . Testing for the other rows and columns (it might be smaller than the other two squares!), we find that the smallest $x$ can be is indeed $8$ ; the other two squares are less than $8$ . Therefore, the answer is $\boxed{8}$",8400113,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_21,1,"Steph scored $15$ baskets out of $20$ attempts in the first half of a game, and $10$ baskets out of $10$ attempts in the second half. Candace took $12$ attempts in the first half and $18$ attempts in the second. In each half, Steph scored a higher percentage of baskets than Candace. Surprisingly they ended with the same overall percentage of baskets scored. How many more baskets did Candace score in the second half than in the first? [asy] size(7cm); draw((-8,27)--(72,27)); draw((16,0)--(16,35)); draw((40,0)--(40,35)); label(""12"", (28,3)); draw((25,6.5)--(25,12)--(31,12)--(31,6.5)--cycle); draw((25,5.5)--(31,5.5)); label(""18"", (56,3)); draw((53,6.5)--(53,12)--(59,12)--(59,6.5)--cycle); draw((53,5.5)--(59,5.5)); draw((53,5.5)--(59,5.5)); label(""20"", (28,18)); label(""15"", (28,24)); draw((25,21)--(31,21)); label(""10"", (56,18)); label(""10"", (56,24)); draw((53,21)--(59,21)); label(""First Half"", (28,31)); label(""Second Half"", (56,31)); label(""Candace"", (2.35,6)); label(""Steph"", (0,21)); [/asy] $\textbf{(A) } 7\qquad\textbf{(B) } 8\qquad\textbf{(C) } 9\qquad\textbf{(D) } 10\qquad\textbf{(E) } 11$","Let $x$ be the number of shots that Candace made in the first half, and let $y$ be the number of shots Candace made in the second half. Since Candace and Steph took the same number of attempts, with an equal percentage of baskets scored, we have $x+y=10+15=25.$ In addition, we have the following inequalities: \[\frac{x}{12}<\frac{15}{20} \implies x<9,\] and \[\frac{y}{18}<\frac{10}{10} \implies y<18.\] Pairing this up with $x+y=25$ we see the only possible solution is $(x,y)=(8,17),$ for an answer of $17-8 = \boxed{9}.$",9401114,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_21,2,"Steph scored $15$ baskets out of $20$ attempts in the first half of a game, and $10$ baskets out of $10$ attempts in the second half. Candace took $12$ attempts in the first half and $18$ attempts in the second. In each half, Steph scored a higher percentage of baskets than Candace. Surprisingly they ended with the same overall percentage of baskets scored. How many more baskets did Candace score in the second half than in the first? [asy] size(7cm); draw((-8,27)--(72,27)); draw((16,0)--(16,35)); draw((40,0)--(40,35)); label(""12"", (28,3)); draw((25,6.5)--(25,12)--(31,12)--(31,6.5)--cycle); draw((25,5.5)--(31,5.5)); label(""18"", (56,3)); draw((53,6.5)--(53,12)--(59,12)--(59,6.5)--cycle); draw((53,5.5)--(59,5.5)); draw((53,5.5)--(59,5.5)); label(""20"", (28,18)); label(""15"", (28,24)); draw((25,21)--(31,21)); label(""10"", (56,18)); label(""10"", (56,24)); draw((53,21)--(59,21)); label(""First Half"", (28,31)); label(""Second Half"", (56,31)); label(""Candace"", (2.35,6)); label(""Steph"", (0,21)); [/asy] $\textbf{(A) } 7\qquad\textbf{(B) } 8\qquad\textbf{(C) } 9\qquad\textbf{(D) } 10\qquad\textbf{(E) } 11$","Clearly, Steph made $15 + 10 = 25$ shots in total. Also, due to parity reasons, the difference between the amount of shots Candace made in the first and second half must be odd. Thus, we can just test 7, 9, and 11, and after doing so we find that the answer is $\boxed{9}.$",9402115,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_21,3,"Steph scored $15$ baskets out of $20$ attempts in the first half of a game, and $10$ baskets out of $10$ attempts in the second half. Candace took $12$ attempts in the first half and $18$ attempts in the second. In each half, Steph scored a higher percentage of baskets than Candace. Surprisingly they ended with the same overall percentage of baskets scored. How many more baskets did Candace score in the second half than in the first? [asy] size(7cm); draw((-8,27)--(72,27)); draw((16,0)--(16,35)); draw((40,0)--(40,35)); label(""12"", (28,3)); draw((25,6.5)--(25,12)--(31,12)--(31,6.5)--cycle); draw((25,5.5)--(31,5.5)); label(""18"", (56,3)); draw((53,6.5)--(53,12)--(59,12)--(59,6.5)--cycle); draw((53,5.5)--(59,5.5)); draw((53,5.5)--(59,5.5)); label(""20"", (28,18)); label(""15"", (28,24)); draw((25,21)--(31,21)); label(""10"", (56,18)); label(""10"", (56,24)); draw((53,21)--(59,21)); label(""First Half"", (28,31)); label(""Second Half"", (56,31)); label(""Candace"", (2.35,6)); label(""Steph"", (0,21)); [/asy] $\textbf{(A) } 7\qquad\textbf{(B) } 8\qquad\textbf{(C) } 9\qquad\textbf{(D) } 10\qquad\textbf{(E) } 11$","Steph made 75 percent of his shots in the first half. He makes all of his shots in the second half. The most baskets Candace could have made in the first half is 8 baskets. The most she could have made in the second half is 17 baskets. Steph makes 25 and misses 5 baskets and the only way for Candace to make 25 shots is to make 8 in the first half and 17 in the second. Thus, $17 - 8 = \boxed{9}.$",9403116,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_22,1,"A bus takes $2$ minutes to drive from one stop to the next, and waits $1$ minute at each stop to let passengers board. Zia takes $5$ minutes to walk from one bus stop to the next. As Zia reaches a bus stop, if the bus is at the previous stop or has already left the previous stop, then she will wait for the bus. Otherwise she will start walking toward the next stop. Suppose the bus and Zia start at the same time toward the library, with the bus $3$ stops behind. After how many minutes will Zia board the bus?404$\textbf{(A) } 17 \qquad \textbf{(B) } 19 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 23$","Initially, suppose that the bus is at Stop $0$ (starting point) and Zia is at Stop $3.$405We construct the following table of $5$ -minute intervals: \[\begin{array}{c||c|c} & & \\ [-2.5ex] \textbf{Time} & \textbf{Bus's Location} & \textbf{Zia's Location} \\ [0.5ex] \hline & & \\ [-2ex] \boldsymbol{5} \ \textbf{Minutes} & \text{Stop} \ 2 \ \text{(Waiting)} & \text{Stop} \ 4 \\ \boldsymbol{10} \ \textbf{Minutes} & \text{Midpoint of Stops} \ 3 \ \text{and} \ 4 & \text{Stop} \ 5 \\ \boldsymbol{15} \ \textbf{Minutes} & \text{Stop} \ 5 \ \text{(Leaving)} & \text{Stop} \ 6 \end{array}\] Note that Zia will wait for the bus after $15$ minutes, and the bus will arrive $2$ minutes later.406Therefore, the answer is $15+2=\boxed{17}.$",17407117,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_22,2,"A bus takes $2$ minutes to drive from one stop to the next, and waits $1$ minute at each stop to let passengers board. Zia takes $5$ minutes to walk from one bus stop to the next. As Zia reaches a bus stop, if the bus is at the previous stop or has already left the previous stop, then she will wait for the bus. Otherwise she will start walking toward the next stop. Suppose the bus and Zia start at the same time toward the library, with the bus $3$ stops behind. After how many minutes will Zia board the bus?408$\textbf{(A) } 17 \qquad \textbf{(B) } 19 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 23$","Since Zia will wait for the bus if the bus is at the previous stop, we can create an equation to solve for when the bus is at the previous stop. The bus travels $\frac{1}{3}$ of a stop per minute, and Zia travels $\frac{1}{5}$ of a stop per minute. Now we create the equation, $\frac{1}{3}m = \frac{1}{5}m + 3 - 1$ (the $-1$ accounts for us wanting to find when the bus reaches the stop before Zia's). Solving, we find that $m=15.$ Now Zia has to wait $2$ minutes for the bus to reach her, so our answer is $15+2=\boxed{17}.$",17409118,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_23,1,"$\triangle$ or $\bigcirc$ is placed in each of the nine squares in a $3$ -by- $3$ grid. Shown below is a sample configuration with three $\triangle$ s in a line. [asy] //diagram size(5cm); defaultpen(linewidth(1.5)); real r = 0.37; path equi = r * dir(-30) -- (r+0.03) * dir(90) -- r * dir(210) -- cycle; draw((0,0)--(0,3)--(3,3)--(3,0)--cycle); draw((0,1)--(3,1)--(3,2)--(0,2)--cycle); draw((1,0)--(1,3)--(2,3)--(2,0)--cycle); draw(circle((3/2,5/2),1/3)); draw(circle((5/2,1/2),1/3)); draw(circle((3/2,3/2),1/3)); draw(shift(0.5,0.38) * equi); draw(shift(1.5,0.38) * equi); draw(shift(0.5,1.38) * equi); draw(shift(2.5,1.38) * equi); draw(shift(0.5,2.38) * equi); draw(shift(2.5,2.38) * equi); [/asy] How many configurations will have three $\triangle$ s in a line and three $\bigcirc$ s in a line?410$\textbf{(A) } 39 \qquad \textbf{(B) } 42 \qquad \textbf{(C) } 78 \qquad \textbf{(D) } 84 \qquad \textbf{(E) } 96$","Notice that diagonals and a vertical-horizontal pair can never work, so the only possibilities are if all lines are vertical or if all lines are horizontal. These are essentially the same, so we'll count up how many work with all lines of shapes vertical, and then multiply by 2 at the end.411We take casework:412Case 1: 3 lines :413In this case, the lines would need to be $2$ of one shape and $1$ of another, so there are $\frac{3!}{2} = 3$ ways to arrange the lines and $2$ ways to pick which shape has only one line. In total, this is $3\cdot 2 = 6.$414Case 2: 2 lines :415In this case, the lines would be one line of triangles, one line of circles, and the last one can be anything that includes both shapes. There are $3! = 6$ ways to arrange the lines and $2^3-2 = 6$ ways to choose the last line. (We subtract $2$ from the last line because one arrangement of the last line is all triangles and the other arrangement of the last line is all circles, which causes Case 2 to overlap with Case 1 and further complicating the solution.) In total, this is $6\cdot 6 = 36.$416Finally, we add and multiply: $2(36+6)=2(42)=\boxed{84}$",84417119,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_23,2,"$\triangle$ or $\bigcirc$ is placed in each of the nine squares in a $3$ -by- $3$ grid. Shown below is a sample configuration with three $\triangle$ s in a line. [asy] //diagram size(5cm); defaultpen(linewidth(1.5)); real r = 0.37; path equi = r * dir(-30) -- (r+0.03) * dir(90) -- r * dir(210) -- cycle; draw((0,0)--(0,3)--(3,3)--(3,0)--cycle); draw((0,1)--(3,1)--(3,2)--(0,2)--cycle); draw((1,0)--(1,3)--(2,3)--(2,0)--cycle); draw(circle((3/2,5/2),1/3)); draw(circle((5/2,1/2),1/3)); draw(circle((3/2,3/2),1/3)); draw(shift(0.5,0.38) * equi); draw(shift(1.5,0.38) * equi); draw(shift(0.5,1.38) * equi); draw(shift(2.5,1.38) * equi); draw(shift(0.5,2.38) * equi); draw(shift(2.5,2.38) * equi); [/asy] How many configurations will have three $\triangle$ s in a line and three $\bigcirc$ s in a line?418$\textbf{(A) } 39 \qquad \textbf{(B) } 42 \qquad \textbf{(C) } 78 \qquad \textbf{(D) } 84 \qquad \textbf{(E) } 96$","We will only consider cases where the three identical symbols are the same column, but at the end we shall double our answer as the same holds true for rows. There are $3$ ways to choose a column with all $\bigcirc$ 's and $2$ ways to choose a column with all $\triangle$ 's. The third column can be filled in $2^3=8$ ways. Therefore, we have a total of $3\cdot2\cdot8=48$ cases. However, we overcounted the cases with $2$ complete columns of with one symbol and $1$ complete column with another symbol. This happens in $2\cdot3=6$ cases. $48-6=42$ . However, we have to remember to double our answer, giving us $\boxed{84}$ ways to complete the grid.",84419120,https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_24,1,"The figure below shows a polygon $ABCDEFGH$ , consisting of rectangles and right triangles. When cut out and folded on the dotted lines, the polygon forms a triangular prism. Suppose that $AH = EF = 8$ and $GH = 14$ . What is the volume of the prism?420[asy] usepackage(""mathptmx""); size(275); defaultpen(linewidth(0.8)); real r = 2, s = 2.5, theta = 14; pair G = (0,0), F = (r,0), C = (r,s), B = (0,s), M = (C+F)/2, I = M + s/2 * dir(-theta); pair N = (B+G)/2, J = N + s/2 * dir(180+theta); pair E = F + r * dir(- 45 - theta/2), D = I+E-F; pair H = J + r * dir(135 + theta/2), A = B+H-J; draw(A--B--C--I--D--E--F--G--J--H--cycle^^rightanglemark(F,I,C)^^rightanglemark(G,J,B)); draw(J--B--G^^C--F--I,linetype (""4 4"")); dot(""$A$"",A,N); dot(""$B$"",B,1.2*N); dot(""$C$"",C,N); dot(""$D$"",D,dir(0)); dot(""$E$"",E,S); dot(""$F$"",F,1.5*dir(-100)); dot(""$G$"",G,S); dot(""$H$"",H,W); dot(""$I$"",I,NE); dot(""$J$"",J,1.5*S); [/asy]421$\textbf{(A)} ~112\qquad\textbf{(B)} ~128\qquad\textbf{(C)} ~192\qquad\textbf{(D)} ~240\qquad\textbf{(E)} ~288$","While imagining the folding, $\overline{AB}$ goes on $\overline{BC},$ $\overline{AH}$ goes on $\overline{CI},$ and $\overline{EF}$ goes on $\overline{FG}.$ So, $BJ=CI=8$ and $FG=BC=8.$ Also, $\overline{HJ}$ becomes an edge parallel to $\overline{FG},$ so that means $HJ=8.$422Since $GH=14,$ then $JG=14-8=6.$ So, the area of $\triangle BJG$ is $\frac{8\cdot6}{2}=24.$ If we let $\triangle BJG$ be the base, then the height is $FG=8.$ So, the volume is $24\cdot8=\boxed{192}.$",192423121,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_1,1,"Luka is making lemonade to sell at a school fundraiser. His recipe requires $4$ times as much water as sugar and twice as much sugar as lemon juice. He uses $3$ cups of lemon juice. How many cups of water does he need?424$\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24$","We have $\text{water} : \text{sugar} : \text{lemon juice} = 4\cdot 2 : 2 : 1 = 8 : 2 : 1,$ so Luka needs $3 \cdot 8 = \boxed{24}$ cups.",24425122,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_1,2,"Luka is making lemonade to sell at a school fundraiser. His recipe requires $4$ times as much water as sugar and twice as much sugar as lemon juice. He uses $3$ cups of lemon juice. How many cups of water does he need?426$\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24$","Since the amount of sugar is twice the amount of lemon juice, Luka uses $3\cdot2=6$ cups of sugar.427Since the amount of water is $4$ times the amount of sugar, he uses $6\cdot4=\boxed{24}$ cups of water.",24428123,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_1,3,"Luka is making lemonade to sell at a school fundraiser. His recipe requires $4$ times as much water as sugar and twice as much sugar as lemon juice. He uses $3$ cups of lemon juice. How many cups of water does he need?429$\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24$","The ratio is $\text{Water}:\text{Sugar}:\text{Lemon Juice},$ or $8:2:1.$ Since we know that Luka used 3 cups of lemon juice, he needs $3\cdot2=6$ cups of sugar. Because the amount of water is $4$ times the amount of sugar Luka needs, he will need $6\cdot4=\boxed{24}$ cups of water.",24430124,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_2,1,"Four friends do yardwork for their neighbors over the weekend, earning $$15, $20, $25,$ and $$40,$ respectively. They decide to split their earnings equally among themselves. In total, how much will the friend who earned $$40$ give to the others?431$\textbf{(A) }$5 \qquad \textbf{(B) }$10 \qquad \textbf{(C) }$15 \qquad \textbf{(D) }$20 \qquad \textbf{(E) }$25$","The friends earn $$\left(15+20+25+40\right)=$100$ in total. Since they decided to split their earnings equally, it follows that each person will get $$\left(\frac{100}{4}\right)=$25$ . Since the friend who earned $$40$ will need to leave with $$25$ , he will have to give $$\left(40-25\right)=\boxed{15}$ to the others.",15432125,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_3,1,"Carrie has a rectangular garden that measures $6$ feet by $8$ feet. She plants the entire garden with strawberry plants. Carrie is able to plant $4$ strawberry plants per square foot, and she harvests an average of $10$ strawberries per plant. How many strawberries can she expect to harvest?433$\textbf{(A) }560 \qquad \textbf{(B) }960 \qquad \textbf{(C) }1120 \qquad \textbf{(D) }1920 \qquad \textbf{(E) }3840$","Note that the unit of the answer is strawberries , which is the product of434By conversion factors, we have \[\left(6 \ \color{red}\cancel{\mathrm{ft}}\color{black}\cdot8 \ \color{red}\cancel{\mathrm{ft}}\color{black}\right)\cdot\left(4 \ \frac{\color{green}\cancel{\mathrm{plants}}}{\color{red}\cancel{\mathrm{ft}^2}}\right)\cdot\left(10 \ \frac{\mathrm{strawberries}}{\color{green}\cancel{\mathrm{plant}}}\right)=6\cdot8\cdot4\cdot10 \ \mathrm{strawberries}=\boxed{1920}.\]",920435126,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_3,2,"Carrie has a rectangular garden that measures $6$ feet by $8$ feet. She plants the entire garden with strawberry plants. Carrie is able to plant $4$ strawberry plants per square foot, and she harvests an average of $10$ strawberries per plant. How many strawberries can she expect to harvest?436$\textbf{(A) }560 \qquad \textbf{(B) }960 \qquad \textbf{(C) }1120 \qquad \textbf{(D) }1920 \qquad \textbf{(E) }3840$","The area of the garden is $6 \cdot 8 = 48$ square feet. Since Carrie plants $4$ strawberry plants per square foot, there are a total of $48 \cdot 4=192$ strawberry plants, each of which produces $10$ strawberries on average. Accordingly, she can expect to harvest $192 \cdot 10 = \boxed{1920}$ strawberries.",920437127,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_4,1,"Three hexagons of increasing size are shown below. Suppose the dot pattern continues so that each successive hexagon contains one more band of dots. How many dots are in the next hexagon?438[asy] // diagram by SirCalcsALot, edited by MRENTHUSIASM size(250); path p = scale(0.8)*unitcircle; pair[] A; pen grey1 = rgb(100/256, 100/256, 100/256); pen grey2 = rgb(183/256, 183/256, 183/256); for (int i=0; i<7; ++i) { A[i] = rotate(60*i)*(1,0);} path hex = A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--cycle; fill(p,grey1); draw(scale(1.25)*hex,black+linewidth(1.25)); pair S = 6A[0]+2A[1]; fill(shift(S)*p,grey1); for (int i=0; i<6; ++i) { fill(shift(S+2*A[i])*p,grey2);} draw(shift(S)*scale(3.25)*hex,black+linewidth(1.25)); pair T = 16A[0]+4A[1]; fill(shift(T)*p,grey1); for (int i=0; i<6; ++i) {   fill(shift(T+2*A[i])*p,grey2);  fill(shift(T+4*A[i])*p,grey1);  fill(shift(T+2*A[i]+2*A[i+1])*p,grey1); } draw(shift(T)*scale(5.25)*hex,black+linewidth(1.25)); [/asy]439$\textbf{(A) }35 \qquad \textbf{(B) }37 \qquad \textbf{(C) }39 \qquad \textbf{(D) }43 \qquad \textbf{(E) }49$","Looking at the rows of each hexagon, we see that the first hexagon has $1$ dot, the second has $2+3+2$ dots, and the third has $3+4+5+4+3$ dots. Given the way the hexagons are constructed, it is clear that this pattern continues. Hence, the fourth hexagon has $4+5+6+7+6+5+4=\boxed{37}$ dots.",37440128,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_4,2,"Three hexagons of increasing size are shown below. Suppose the dot pattern continues so that each successive hexagon contains one more band of dots. How many dots are in the next hexagon?441[asy] // diagram by SirCalcsALot, edited by MRENTHUSIASM size(250); path p = scale(0.8)*unitcircle; pair[] A; pen grey1 = rgb(100/256, 100/256, 100/256); pen grey2 = rgb(183/256, 183/256, 183/256); for (int i=0; i<7; ++i) { A[i] = rotate(60*i)*(1,0);} path hex = A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--cycle; fill(p,grey1); draw(scale(1.25)*hex,black+linewidth(1.25)); pair S = 6A[0]+2A[1]; fill(shift(S)*p,grey1); for (int i=0; i<6; ++i) { fill(shift(S+2*A[i])*p,grey2);} draw(shift(S)*scale(3.25)*hex,black+linewidth(1.25)); pair T = 16A[0]+4A[1]; fill(shift(T)*p,grey1); for (int i=0; i<6; ++i) {   fill(shift(T+2*A[i])*p,grey2);  fill(shift(T+4*A[i])*p,grey1);  fill(shift(T+2*A[i]+2*A[i+1])*p,grey1); } draw(shift(T)*scale(5.25)*hex,black+linewidth(1.25)); [/asy]442$\textbf{(A) }35 \qquad \textbf{(B) }37 \qquad \textbf{(C) }39 \qquad \textbf{(D) }43 \qquad \textbf{(E) }49$","The dots in the next hexagon have four bands. From innermost to outermost:443Together, the answer is $1+6+12+18=\boxed{37}.$",37444129,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_4,3,"Three hexagons of increasing size are shown below. Suppose the dot pattern continues so that each successive hexagon contains one more band of dots. How many dots are in the next hexagon?445[asy] // diagram by SirCalcsALot, edited by MRENTHUSIASM size(250); path p = scale(0.8)*unitcircle; pair[] A; pen grey1 = rgb(100/256, 100/256, 100/256); pen grey2 = rgb(183/256, 183/256, 183/256); for (int i=0; i<7; ++i) { A[i] = rotate(60*i)*(1,0);} path hex = A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--cycle; fill(p,grey1); draw(scale(1.25)*hex,black+linewidth(1.25)); pair S = 6A[0]+2A[1]; fill(shift(S)*p,grey1); for (int i=0; i<6; ++i) { fill(shift(S+2*A[i])*p,grey2);} draw(shift(S)*scale(3.25)*hex,black+linewidth(1.25)); pair T = 16A[0]+4A[1]; fill(shift(T)*p,grey1); for (int i=0; i<6; ++i) {   fill(shift(T+2*A[i])*p,grey2);  fill(shift(T+4*A[i])*p,grey1);  fill(shift(T+2*A[i]+2*A[i+1])*p,grey1); } draw(shift(T)*scale(5.25)*hex,black+linewidth(1.25)); [/asy]446$\textbf{(A) }35 \qquad \textbf{(B) }37 \qquad \textbf{(C) }39 \qquad \textbf{(D) }43 \qquad \textbf{(E) }49$","The first hexagon has $1$ dot, the second hexagon has $1+6$ dots, the third hexagon has $1+6+12$ dots, and so on. The pattern continues since to go from hexagon $n$ to hexagon $(n+1),$ we add a new band of dots around the outside of the existing ones, with each side of the band having side length $(n+1).$ Thus, the number of dots added is $6(n+1)-6 = 6n$ (we subtract $6$ as each of the corner hexagons in the band is counted as part of two sides.). We therefore predict that the fourth hexagon has $1+6+12+18=\boxed{37}$ dots.",37447130,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_5,1,"Three fourths of a pitcher is filled with pineapple juice. The pitcher is emptied by pouring an equal amount of juice into each of $5$ cups. What percent of the total capacity of the pitcher did each cup receive?448$\textbf{(A) }5 \qquad \textbf{(B) }10 \qquad \textbf{(C) }15 \qquad \textbf{(D) }20 \qquad \textbf{(E) }25$","Each cup is filled with $\frac{3}{4} \cdot \frac{1}{5} = /frac{3}{20}$ of the amount of juice in the pitcher, so the percentage is $\frac{3}{20} \cdot 100 = \boxed{15}$",15449131,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_5,2,"Three fourths of a pitcher is filled with pineapple juice. The pitcher is emptied by pouring an equal amount of juice into each of $5$ cups. What percent of the total capacity of the pitcher did each cup receive?450$\textbf{(A) }5 \qquad \textbf{(B) }10 \qquad \textbf{(C) }15 \qquad \textbf{(D) }20 \qquad \textbf{(E) }25$","The pitcher is $\frac{3}{4}$ full, i.e. $75\%$ full. Therefore each cup receives $\frac{75}{5}=\boxed{15}$ percent of the total capacity.",15451132,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_5,3,"Three fourths of a pitcher is filled with pineapple juice. The pitcher is emptied by pouring an equal amount of juice into each of $5$ cups. What percent of the total capacity of the pitcher did each cup receive?452$\textbf{(A) }5 \qquad \textbf{(B) }10 \qquad \textbf{(C) }15 \qquad \textbf{(D) }20 \qquad \textbf{(E) }25$","Assume that the pitcher has a total capacity of $100$ ounces. Since it is filled three fourths with pineapple juice, it contains $75$ ounces of pineapple juice, which means that each cup will contain $\frac{75}{5}=15$ ounces of pineapple juice. Since the total capacity of the pitcher was $100$ ounces, it follows that each cup received $15\%$ of the total capacity of the pitcher, yielding $\boxed{15}$ as the answer.",15453133,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_7,1,"How many integers between $2020$ and $2400$ have four distinct digits arranged in increasing order? (For example, $2347$ is one integer.)454$\textbf{(A) }\text{9} \qquad \textbf{(B) }\text{10} \qquad \textbf{(C) }\text{15} \qquad \textbf{(D) }\text{21}\qquad \textbf{(E) }\text{28}$","Firstly, observe that the second digit of such a number cannot be $1$ or $2$ , because the digits must be distinct and increasing. The second digit also cannot be $4$ as the number must be less than $2400$ , so it must be $3$ . It remains to choose the latter two digits, which must be $2$ distinct digits from $\left\{4,5,6,7,8,9\right\}$ . That can be done in $\binom{6}{2} = \frac{6 \cdot 5}{2 \cdot 1} = 15$ ways; there is then only $1$ way to order the digits, namely in increasing order. This means the answer is $\boxed{15}$",15455134,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_7,2,"How many integers between $2020$ and $2400$ have four distinct digits arranged in increasing order? (For example, $2347$ is one integer.)456$\textbf{(A) }\text{9} \qquad \textbf{(B) }\text{10} \qquad \textbf{(C) }\text{15} \qquad \textbf{(D) }\text{21}\qquad \textbf{(E) }\text{28}$","As in Solution 1, we find that the first two digits must be $23$ , and the third digit must be at least $4$ . If it is $4$ , then there are $5$ choices for the last digit, namely $5$ $6$ $7$ $8$ , or $9$ . Similarly, if the third digit is $5$ , there are $4$ choices for the last digit, namely $6$ $7$ $8$ , and $9$ ; if $6$ , there are $3$ choices; if $7$ , there are $2$ choices; and if $8$ , there is $1$ choice. It follows that the total number of such integers is $5+4+3+2+1=\boxed{15}$",15457135,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_8,1,"Ricardo has $2020$ coins, some of which are pennies ( $1$ -cent coins) and the rest of which are nickels ( $5$ -cent coins). He has at least one penny and at least one nickel. What is the difference in cents between the greatest possible and least amounts of money that Ricardo can have?458$\textbf{(A) }\text{806} \qquad \textbf{(B) }\text{8068} \qquad \textbf{(C) }\text{8072} \qquad \textbf{(D) }\text{8076}\qquad \textbf{(E) }\text{8082}$","Clearly, the amount of money Ricardo has will be maximized when he has the maximum number of nickels. Since he must have at least one penny, the greatest number of nickels he can have is $2019$ , giving a total of $(2019\cdot 5 + 1)$ cents. Analogously, the amount of money he has will be least when he has the greatest number of pennies; as he must have at least one nickel, the greatest number of pennies he can have is also $2019$ , giving him a total of $(2019\cdot 1 + 5)$ cents. Hence the required difference is \[(2019\cdot 5 + 1)-(2019\cdot 1 + 5)=2019\cdot 4-4=4\cdot 2018=\boxed{8072}\]",72459136,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_8,2,"Ricardo has $2020$ coins, some of which are pennies ( $1$ -cent coins) and the rest of which are nickels ( $5$ -cent coins). He has at least one penny and at least one nickel. What is the difference in cents between the greatest possible and least amounts of money that Ricardo can have?460$\textbf{(A) }\text{806} \qquad \textbf{(B) }\text{8068} \qquad \textbf{(C) }\text{8072} \qquad \textbf{(D) }\text{8076}\qquad \textbf{(E) }\text{8082}$","Suppose Ricardo has $p$ pennies, so then he has $(2020-p)$ nickels. In order to have at least one penny and at least one nickel, we require $p \geq 1$ and $2020 - p \geq 1$ , i.e. $1 \leq p \leq 2019$ . The number of cents he has is $p+5(2020-p) = 10100-4p$ , so the maximum is $10100-4 \cdot 1$ and the minimum is $10100 - 4 \cdot 2019$ , and the difference is therefore \[(10100 - 4\cdot 1) - (10100 - 4\cdot 2019) = 4\cdot 2019 - 4 = 4\cdot 2018 = \boxed{8072}\]",72461137,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_10,1,"Zara has a collection of $4$ marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?462$\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24$","Let the Aggie, Bumblebee, Steelie, and Tiger, be referred to by $A,B,S,$ and $T$ , respectively. If we ignore the constraint that $S$ and $T$ cannot be next to each other, we get a total of $4!=24$ ways to arrange the 4 marbles. We now simply have to subtract out the number of ways that $S$ and $T$ can be next to each other. If we place $S$ and $T$ next to each other in that order, then there are three places that we can place them, namely in the first two slots, in the second two slots, or in the last two slots (i.e. $ST\square\square, \square ST\square, \square\square ST$ ). However, we could also have placed $S$ and $T$ in the opposite order (i.e. $TS\square\square, \square TS\square, \square\square TS$ ). Thus there are 6 ways of placing $S$ and $T$ directly next to each other. Next, notice that for each of these placements, we have two open slots for placing $A$ and $B$ . Specifically, we can place $A$ in the first open slot and $B$ in the second open slot or switch their order and place $B$ in the first open slot and $A$ in the second open slot. This gives us a total of $6\times 2=12$ ways to place $S$ and $T$ next to each other. Subtracting this from the total number of arrangements gives us $24-12=12$ total arrangements $\implies\boxed{12}$",12463138,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_10,2,"Zara has a collection of $4$ marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?464$\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24$","Let's try complementary counting. There $4!$ ways to arrange the 4 marbles. However, there are $2\cdot3!$ arrangements where Steelie and Tiger are next to each other. (Think about permutations of the element ST, A, and B or TS, A, and B). Thus, \[4!-2\cdot3!=\boxed{12}\]",12465139,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_10,3,"Zara has a collection of $4$ marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?466$\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24$","We use complementary counting: we will count the numbers of ways where Steelie and Tiger are together and subtract that from the total count. Treat the Steelie and the Tiger as a ""super marble."" There are $2!$ ways to arrange Steelie and Tiger within this ""super marble."" Then there are $3!$ ways to arrange the ""super marble"" and Zara's two other marbles in a row. Since there are $4!$ ways to arrange the marbles without any restrictions, the answer is given by $4!-2!\cdot 3!=\boxed{12}$",12467140,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_10,4,"Zara has a collection of $4$ marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?468$\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24$","We will use the following469$\textbf{Georgeooga-Harryooga Theorem:}$ The Georgeooga-Harryooga Theorem states that if you have $a$ distinguishable objects and $b$ of them cannot be together, then there are $\frac{(a-b)!(a-b+1)!}{b!}$ ways to arrange the objects.470$\textit{Proof. (Created by AoPS user RedFireTruck)}$471Let our group of $a$ objects be represented like so $1$ $2$ $3$ , ..., $a-1$ $a$ . Let the last $b$ objects be the ones we can't have together.472Then we can organize our objects like so $\square1\square2\square3\square...\square a-b-1\square a-b\square$473We have $(a-b)!$ ways to arrange the objects in that list.474Now we have $a-b+1$ blanks and $b$ other objects so we have $_{a-b+1}P_{b}=\frac{(a-b+1)!}{(a-2b+1)!}$ ways to arrange the objects we can't put together.475By fundamental counting principle our answer is $\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}$476Proof by RedFireTruck talk ) 12:09, 1 February 2021 (EST)477Back to the problem. By the Georgeooga-Harryooga Theorem , our answer is $\frac{(4-2)!(4-2+1)!}{(4-2\cdot2+1)!}=\boxed{12}$",12478141,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_11,1,"After school, Maya and Naomi headed to the beach, $6$ miles away. Maya decided to bike while Naomi took a bus. The graph below shows their journeys, indicating the time and distance traveled. What was the difference, in miles per hour, between Naomi's and Maya's average speeds?479[asy] unitsize(1.25cm); dotfactor = 10; pen shortdashed=linetype(new real[] {2.7,2.7});  for (int i = 0; i < 6; ++i) {     for (int j = 0; j < 6; ++j) {         draw((i,0)--(i,6), grey);         draw((0,j)--(6,j), grey);     } }  for (int i = 1; i <= 6; ++i) {     draw((-0.1,i)--(0.1,i),linewidth(1.25));     draw((i,-0.1)--(i,0.1),linewidth(1.25));     label(string(5*i), (i,0), 2*S);     label(string(i), (0, i), 2*W);  }  draw((0,0)--(0,6)--(6,6)--(6,0)--(0,0)--cycle,linewidth(1.25));  label(rotate(90) * ""Distance (miles)"", (-0.5,3), W); label(""Time (minutes)"", (3,-0.5), S);  dot(""Naomi"", (2,6), 3*dir(305)); dot((6,6));  label(""Maya"", (4.45,3.5));  draw((0,0)--(1.15,1.3)--(1.55,1.3)--(3.15,3.2)--(3.65,3.2)--(5.2,5.2)--(5.4,5.2)--(6,6),linewidth(1.35)); draw((0,0)--(0.4,0.1)--(1.15,3.7)--(1.6,3.7)--(2,6),linewidth(1.35)+shortdashed); [/asy]480$\textbf{(A) }6 \qquad \textbf{(B) }12 \qquad \textbf{(C) }18 \qquad \textbf{(D) }20 \qquad \textbf{(E) }24$","Naomi travels $6$ miles in a time of $10$ minutes, which is equivalent to $\dfrac{1}{6}$ of an hour. Since $\text{speed} = \frac{\text{distance}}{\text{time}}$ , her speed is $\frac{6}{\left(\frac{1}{6}\right)} = 36$ mph. By a similar calculation, Maya's speed is $12$ mph, so the answer is $36-12 = \boxed{24}$",24481142,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_11,2,"After school, Maya and Naomi headed to the beach, $6$ miles away. Maya decided to bike while Naomi took a bus. The graph below shows their journeys, indicating the time and distance traveled. What was the difference, in miles per hour, between Naomi's and Maya's average speeds?482[asy] unitsize(1.25cm); dotfactor = 10; pen shortdashed=linetype(new real[] {2.7,2.7});  for (int i = 0; i < 6; ++i) {     for (int j = 0; j < 6; ++j) {         draw((i,0)--(i,6), grey);         draw((0,j)--(6,j), grey);     } }  for (int i = 1; i <= 6; ++i) {     draw((-0.1,i)--(0.1,i),linewidth(1.25));     draw((i,-0.1)--(i,0.1),linewidth(1.25));     label(string(5*i), (i,0), 2*S);     label(string(i), (0, i), 2*W);  }  draw((0,0)--(0,6)--(6,6)--(6,0)--(0,0)--cycle,linewidth(1.25));  label(rotate(90) * ""Distance (miles)"", (-0.5,3), W); label(""Time (minutes)"", (3,-0.5), S);  dot(""Naomi"", (2,6), 3*dir(305)); dot((6,6));  label(""Maya"", (4.45,3.5));  draw((0,0)--(1.15,1.3)--(1.55,1.3)--(3.15,3.2)--(3.65,3.2)--(5.2,5.2)--(5.4,5.2)--(6,6),linewidth(1.35)); draw((0,0)--(0.4,0.1)--(1.15,3.7)--(1.6,3.7)--(2,6),linewidth(1.35)+shortdashed); [/asy]483$\textbf{(A) }6 \qquad \textbf{(B) }12 \qquad \textbf{(C) }18 \qquad \textbf{(D) }20 \qquad \textbf{(E) }24$","Naomi's speed of $6$ miles in $10$ minutes is equivalent to $6 \cdot 6 = 36$ miles per hour, while Maya's speed of $6$ miles in $30$ minutes (i.e. half an hour) is equivalent to $6 \cdot 2 = 12$ miles per hour. The difference is consequently $36-12=\boxed{24}$",24484143,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_12,1,"For a positive integer $n$ , the factorial notation $n!$ represents the product of the integers from $n$ to $1$ . What value of $N$ satisfies the following equation? \[5!\cdot 9!=12\cdot N!\]485$\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14\qquad$","We have $5! = 2 \cdot 3 \cdot 4 \cdot 5$ , and $2 \cdot 5 \cdot 9! = 10 \cdot 9! = 10!$ . Therefore, the equation becomes $3 \cdot 4 \cdot 10! = 12 \cdot N!$ , and so $12 \cdot 10! = 12 \cdot N!$ . Cancelling the $12$ s, it is clear that $N=\boxed{10}$",10486144,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_12,2,"For a positive integer $n$ , the factorial notation $n!$ represents the product of the integers from $n$ to $1$ . What value of $N$ satisfies the following equation? \[5!\cdot 9!=12\cdot N!\]487$\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14\qquad$","Since $5! = 120$ , we obtain $120\cdot 9!=12\cdot N!$ , which becomes $12\cdot 10\cdot 9!=12\cdot N!$ and thus $12 \cdot 10!=12\cdot N!$ . We therefore deduce $N=\boxed{10}$",10488145,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_12,3,"For a positive integer $n$ , the factorial notation $n!$ represents the product of the integers from $n$ to $1$ . What value of $N$ satisfies the following equation? \[5!\cdot 9!=12\cdot N!\]489$\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14\qquad$","We notice that $5! \cdot 9! = (5!)^2 \cdot (9 \cdot 8 \cdot 7 \cdot 6).$490We know that $5! = 120,$ so we have $120(5! \cdot 9 \cdot 8 \cdot 7 \cdot 6) = 12 \cdot N!$491Isolating $N!$ we have $N! = 10 \cdot 5! \cdot 9 \cdot 8 \cdot 7 \cdot 6 \Rightarrow N! = 10! \Rightarrow N = \boxed{10}.$",10492146,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_13,1,"Jamal has a drawer containing $6$ green socks, $18$ purple socks, and $12$ orange socks. After adding more purple socks, Jamal noticed that there is now a $60\%$ chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?493$\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24$","After Jamal adds $x$ purple socks, he has $(18+x)$ purple socks and $6+18+12+x=(36+x)$ total socks. This means the probability of drawing a purple sock is $\frac{18+x}{36+x}$ , so we obtain \[\frac{18+x}{36+x}=\frac{3}{5}\] Since $\frac{18+9}{36+9}=\frac{27}{45}=\frac{3}{5}$ , the answer is $\boxed{9}$",9494147,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_13,2,"Jamal has a drawer containing $6$ green socks, $18$ purple socks, and $12$ orange socks. After adding more purple socks, Jamal noticed that there is now a $60\%$ chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?495$\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24$","As in Solution 1, we have the equation $\frac{18+x}{36+x}=\frac{3}{5}$ . Cross-multiplying yields $90+5x=108+3x \Rightarrow 2x=18 \Rightarrow x=9$ . Thus, Jamal added $\boxed{9}$ purple socks.",9496148,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_13,3,"Jamal has a drawer containing $6$ green socks, $18$ purple socks, and $12$ orange socks. After adding more purple socks, Jamal noticed that there is now a $60\%$ chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?497$\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24$","$6$ green socks and $12$ orange socks together should be $100\%-60\% = 40\%$ of the new total number of socks, so that new total must be $\frac{6+12}{0.4}= 45$ . Therefore, $45-6-18-12=\boxed{9}$ purple socks were added.",9498149,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_14,1,"There are $20$ cities in the County of Newton. Their populations are shown in the bar chart below. The average population of all the cities is indicated by the horizontal dashed line. Which of the following is closest to the total population of all $20$ cities?499[asy] // made by SirCalcsALot  size(300);  pen shortdashed=linetype(new real[] {6,6});  for (int i = 2000; i < 9000; i = i + 2000) {     draw((0,i)--(11550,i), linewidth(0.5)+1.5*grey);     label(string(i), (0,i), W); }   for (int i = 500; i < 9300; i=i+500) {     draw((0,i)--(150,i),linewidth(1.25));     if (i % 2000 == 0) {         draw((0,i)--(250,i),linewidth(1.25));     } }  int[] data = {8750, 3800, 5000, 2900, 6400, 7500, 4100, 1400, 2600, 1470, 2600, 7100, 4070, 7500, 7000, 8100, 1900, 1600, 5850, 5750}; int data_length = 20;  int r = 550; for (int i = 0; i < data_length; ++i) {     fill(((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+2)*r-100, data[i])--((i+2)*r-100,0)--cycle, 1.5*grey);     draw(((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+2)*r-100, data[i])--((i+2)*r-100,0)); }  draw((0,4750)--(11450,4750),shortdashed);  label(""Cities"", (11450*0.5,0), S); label(rotate(90)*""Population"", (0,9000*0.5), 10*W);  // axis draw((0,0)--(0,9300), linewidth(1.25)); draw((0,0)--(11550,0), linewidth(1.25)); [/asy]500$\textbf{(A) }65000 \qquad \textbf{(B) }75000 \qquad \textbf{(C) }85000 \qquad \textbf{(D) }95000 \qquad \textbf{(E) }105000$","We can see that the dotted line is exactly halfway between $4500$ and $5000$ , so it is at $4750$ . As this is the average population of all $20$ cities, the total population is simply $4750 \cdot 20 = \boxed{95000}$",0501150,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_15,1,"Suppose $15\%$ of $x$ equals $20\%$ of $y.$ What percentage of $x$ is $y?$502$\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300$","Since $20\% = \frac{1}{5}$ , multiplying the given condition by $5$ shows that $y$ is $15 \cdot 5 = \boxed{75}$ percent of $x$",75503151,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_15,2,"Suppose $15\%$ of $x$ equals $20\%$ of $y.$ What percentage of $x$ is $y?$504$\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300$","Letting $x=100$ (without loss of generality), the condition becomes $0.15\cdot 100 = 0.2\cdot y \Rightarrow 15 = \frac{y}{5} \Rightarrow y=75$ . Clearly, it follows that $y$ is $75\%$ of $x$ , so the answer is $\boxed{75}$",75505152,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_15,3,"Suppose $15\%$ of $x$ equals $20\%$ of $y.$ What percentage of $x$ is $y?$506$\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300$","We have $15\%=\frac{3}{20}$ and $20\%=\frac{1}{5}$ , so $\frac{3}{20}x=\frac{1}{5}y$ . Solving for $y$ , we multiply by $5$ to give $y = \frac{15}{20}x = \frac{3}{4}x$ , so the answer is $\boxed{75}$",75507153,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_15,4,"Suppose $15\%$ of $x$ equals $20\%$ of $y.$ What percentage of $x$ is $y?$508$\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300$","We are given $0.15x = 0.20y$ , so we may assume without loss of generality that $x=20$ and $y=15$ . This means $\frac{y}{x}=\frac{15}{20}=\frac{75}{100}$ , and thus the answer is $\boxed{75}$",75509154,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_15,5,"Suppose $15\%$ of $x$ equals $20\%$ of $y.$ What percentage of $x$ is $y?$510$\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300$","$15\%$ of $x$ is $0.15x$ , and $20\%$ of $y$ is $0.20y$ . We put $0.15x$ and $0.20y$ into an equation, creating $0.15x = 0.20y$ because $0.15x$ equals $0.20y$ . Solving for $y$ , dividing $0.2$ to both sides, we get $y = \frac{15}{20}x = \frac{3}{4}x$ , so the answer is $\boxed{75}$",75511155,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_15,6,"Suppose $15\%$ of $x$ equals $20\%$ of $y.$ What percentage of $x$ is $y?$512$\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300$","$15\%$ of $x$ can be written as $\frac{15}{100}x$ , or $\frac{15x}{100}$ $20\%$ of $y$ can similarly be written as $\frac{20}{100}y$ , or $\frac{20y}{100}$ . So now, $\frac{15x}{100} = \frac{20y}{100}$ . Using cross-multiplication, we can simplify the equation as: $1500x = 2000y$ . Dividing both sides by $500$ , we get: $3x = 4y$ $\frac{3}{4}$ is the same thing as $75\%$ , so the answer is $\boxed{75}$",75513156,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_16,2,"Each of the points $A,B,C,D,E,$ and $F$ in the figure below represents a different digit from $1$ to $6.$ Each of the five lines shown passes through some of these points. The digits along each line are added to produce five sums, one for each line. The total of the five sums is $47.$ What is the digit represented by $B?$514[asy] size(200); dotfactor = 10;  pair p1 = (-28,0); pair p2 = (-111,213); draw(p1--p2,linewidth(1));  pair p3 = (-160,0); pair p4 = (-244,213); draw(p3--p4,linewidth(1));  pair p5 = (-316,0); pair p6 = (-67,213); draw(p5--p6,linewidth(1));  pair p7 = (0, 68); pair p8 = (-350,10); draw(p7--p8,linewidth(1));  pair p9 = (0, 150); pair p10 = (-350, 62); draw(p9--p10,linewidth(1));  pair A = intersectionpoint(p1--p2, p5--p6); dot(""$A$"", A, 2*W);  pair B = intersectionpoint(p5--p6, p3--p4); dot(""$B$"", B, 2*WNW);  pair C = intersectionpoint(p7--p8, p5--p6); dot(""$C$"", C, 1.5*NW);  pair D = intersectionpoint(p3--p4, p7--p8); dot(""$D$"", D, 2*NNE);  pair EE = intersectionpoint(p1--p2, p7--p8); dot(""$E$"", EE, 2*NNE);  pair F = intersectionpoint(p1--p2, p9--p10); dot(""$F$"", F, 2*NNE); [/asy]515$\textbf{(A) }1 \qquad \textbf{(B) }2 \qquad \textbf{(C) }3 \qquad \textbf{(D) }4 \qquad \textbf{(E) }5$","Following the first few steps of Solution 1, we have $2(A+C+D+E+F)+3B=47$ . Because an even number ( $2(A+C+D+E+F)$ ) subtracted from an odd number (47) is always odd, we know that $3B$ is odd, showing that $B$ is odd. Now we know that $B$ is either 1, 3, or 5. If we try $B=1$ , we get $43=47$ which is not true. Testing $B=3$ , we get $45=47$ , which is also not true. Therefore, we have $B = \boxed{5}$",5516157,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_17,2,"How many positive integer factors of $2020$ have more than $3$ factors? (As an example, $12$ has $6$ factors, namely $1,2,3,4,6,$ and $12.$517$\textbf{(A) }6 \qquad \textbf{(B) }7 \qquad \textbf{(C) }8 \qquad \textbf{(D) }9 \qquad \textbf{(E) }10$","As in Solution 1, we prime factorize $2020$ as $2^2\cdot 5\cdot 101$ , and we recall the standard formula that the number of positive factors of an integer is found by adding $1$ to each exponent in its prime factorization, and then multiplying these. Thus $2020$ has $(2+1)(1+1)(1+1) = 12$ factors. The only number which has one factor is $1$ . For a number to have exactly two factors, it must be prime, and the only prime factors of $2020$ are $2$ $5$ , and $101$ . For a number to have three factors, it must be a square of a prime (this follows from the standard formula mentioned above), and from the prime factorization, the only square of a prime that is a factor of $2020$ is $4$ . Thus, there are $5$ factors of $2020$ which themselves have $1$ $2$ , or $3$ factors (namely $1$ $2$ $4$ $5$ , and $101$ ), so the number of factors of $2020$ that have more than $3$ factors is $12-5=\boxed{7}$",7518158,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_17,3,"How many positive integer factors of $2020$ have more than $3$ factors? (As an example, $12$ has $6$ factors, namely $1,2,3,4,6,$ and $12.$519$\textbf{(A) }6 \qquad \textbf{(B) }7 \qquad \textbf{(C) }8 \qquad \textbf{(D) }9 \qquad \textbf{(E) }10$","Let $d(n)$ be the number of factors of n. We know by prime factorization that $d(2020) = 12$ . These $12$ numbers can be divided into unordered pairs ${a,b}$ where $ab = 2020$ . Since $d(2020) = d(a)d(b)$ , one of $d(a), d(b)$ has $3$ or less factors and the other has $4$ or more - in to total $6$ factors of $2020$ with more than $3$ factors. However, this argument has exceptions where $a$ and $b$ share a nontrivial common factor, which in this case can only be two. There are two cases - One in which $5$ and $101$ divide the same factor, WLOG assumed to be $a$ , so that $d(a) = 2^3 > 3$ and $d(b) = 2$ , as otherwise. In the other case, $a = 5\cdot2$ and $b = 101\cdot2$ , so that $d(a) = d(b) = 4$ . This adds one factor with more than $3$ factors, so the answer is $\boxed{7}$",7520159,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_18,1,"Rectangle $ABCD$ is inscribed in a semicircle with diameter $\overline{FE},$ as shown in the figure. Let $DA=16,$ and let $FD=AE=9.$ What is the area of $ABCD?$521[asy]  draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot(""$A$"",(8,0), 1.25*S); dot(""$B$"",(8,15), 1.25*N); dot(""$C$"",(-8,15), 1.25*N); dot(""$D$"",(-8,0), 1.25*S); dot(""$E$"",(17,0), 1.25*S); dot(""$F$"",(-17,0), 1.25*S); label(""$16$"",(0,0),N); label(""$9$"",(12.5,0),N); label(""$9$"",(-12.5,0),N);  [/asy] $\textbf{(A) }240 \qquad \textbf{(B) }248 \qquad \textbf{(C) }256 \qquad \textbf{(D) }264 \qquad \textbf{(E) }272$","[asy]  draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot(""$A$"",(8,0), 1.25*S); dot(""$B$"",(8,15), 1.25*N); dot(""$C$"",(-8,15), 1.25*N); dot(""$D$"",(-8,0), 1.25*S); dot(""$E$"",(17,0), 1.25*S); dot(""$F$"",(-17,0), 1.25*S); label(""$16$"",(0,0),N); label(""$9$"",(12.5,0),N); label(""$9$"",(-12.5,0),N); dot(""$O$"", (0,0), 1.25*S); draw((0,0)--(-8,15));[/asy]522Let $O$ be the center of the semicircle. The diameter of the semicircle is $9+16+9=34$ , so $OC = 17$ . By symmetry, $O$ is the midpoint of $DA$ , so $OD=OA=\frac{16}{2}= 8$ . By the Pythagorean theorem in right-angled triangle $ODC$ (or $OBA$ ), we have that $CD$ (or $AB$ ) is $\sqrt{17^2-8^2}=15$ . Accordingly, the area of $ABCD$ is $16\cdot 15=\boxed{240}$",240523160,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_18,2,"Rectangle $ABCD$ is inscribed in a semicircle with diameter $\overline{FE},$ as shown in the figure. Let $DA=16,$ and let $FD=AE=9.$ What is the area of $ABCD?$524[asy]  draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot(""$A$"",(8,0), 1.25*S); dot(""$B$"",(8,15), 1.25*N); dot(""$C$"",(-8,15), 1.25*N); dot(""$D$"",(-8,0), 1.25*S); dot(""$E$"",(17,0), 1.25*S); dot(""$F$"",(-17,0), 1.25*S); label(""$16$"",(0,0),N); label(""$9$"",(12.5,0),N); label(""$9$"",(-12.5,0),N);  [/asy] $\textbf{(A) }240 \qquad \textbf{(B) }248 \qquad \textbf{(C) }256 \qquad \textbf{(D) }264 \qquad \textbf{(E) }272$","Let the midpoint of segment $FE$ be the origin. Evidently, point $D=(-8,0)$ and $A=(8,0)$ . Since points $C$ and $B$ share $x$ -coordinates with $D$ and $A$ respectively, it suffices to find the $y$ -coordinate of $B$ (which will be the height of the rectangle) and multiply this by $DA$ (which we know is $16$ ). The radius of the semicircle is $\frac{9+16+9}{2} = 17$ , so the whole circle has equation $x^2+y^2=289$ ; as already stated, $B$ has the same $x$ -coordinate as $A$ , i.e. $8$ , so substituting this into the equation shows that $y=\pm15$ . Since $y>0$ at $B$ , the y-coordinate of $B$ is $15$ . Therefore, the answer is $16\cdot 15 = \boxed{240}$",240525161,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_18,4,"Rectangle $ABCD$ is inscribed in a semicircle with diameter $\overline{FE},$ as shown in the figure. Let $DA=16,$ and let $FD=AE=9.$ What is the area of $ABCD?$526[asy]  draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot(""$A$"",(8,0), 1.25*S); dot(""$B$"",(8,15), 1.25*N); dot(""$C$"",(-8,15), 1.25*N); dot(""$D$"",(-8,0), 1.25*S); dot(""$E$"",(17,0), 1.25*S); dot(""$F$"",(-17,0), 1.25*S); label(""$16$"",(0,0),N); label(""$9$"",(12.5,0),N); label(""$9$"",(-12.5,0),N);  [/asy] $\textbf{(A) }240 \qquad \textbf{(B) }248 \qquad \textbf{(C) }256 \qquad \textbf{(D) }264 \qquad \textbf{(E) }272$","Draw the other half of the circle as follows: [asy]  draw(arc((0,0),17,360,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot(""$A$"",(8,0), 1.25*SE); dot(""$B$"",(8,15), 1.25*N); dot(""$C$"",(-8,15), 1.25*N); dot(""$D$"",(-8,0), 1.25*SW); dot(""$E$"",(17,0), 1.25*E); dot(""$F$"",(-17,0), 1.25*W); label(""$16$"",(0,0),N); label(""$9$"",(12.5,0),N); label(""$9$"",(-12.5,0),N); draw((-8,-15)--(-8,0)--(8,0)--(8,-15)--cycle); dot(""$B'$"",(8,-15), 1.25*S); dot(""$C'$"",(-8,-15), 1.25*S); [/asy] By the Power of a Point Theorem $FD\cdot DE = CD\cdot C'D$ . By symmetry, $CD = C'D$ . We see that $FD = 9$ and $DE = 16 + 9 = 25$ . Substituting in these values, $9\cdot 25 = CD^2$ , giving $CD^2 = 225$ and $CD = 15$ .527The area of the rectangle is therefore $15\cdot 16 = \boxed{240}$",240528162,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_18,5,"Rectangle $ABCD$ is inscribed in a semicircle with diameter $\overline{FE},$ as shown in the figure. Let $DA=16,$ and let $FD=AE=9.$ What is the area of $ABCD?$529[asy]  draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot(""$A$"",(8,0), 1.25*S); dot(""$B$"",(8,15), 1.25*N); dot(""$C$"",(-8,15), 1.25*N); dot(""$D$"",(-8,0), 1.25*S); dot(""$E$"",(17,0), 1.25*S); dot(""$F$"",(-17,0), 1.25*S); label(""$16$"",(0,0),N); label(""$9$"",(12.5,0),N); label(""$9$"",(-12.5,0),N);  [/asy] $\textbf{(A) }240 \qquad \textbf{(B) }248 \qquad \textbf{(C) }256 \qquad \textbf{(D) }264 \qquad \textbf{(E) }272$","[asy]  draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot(""$A$"",(8,0), 1.25*S); dot(""$B$"",(8,15), 1.25*N); dot(""$C$"",(-8,15), 1.25*N); dot(""$D$"",(-8,0), 1.25*S); dot(""$E$"",(17,0), 1.25*S); dot(""$F$"",(-17,0), 1.25*S); label(""$16$"",(0,0),S); label(""$9$"",(12.5,0),N); label(""$9$"",(-12.5,0),N); dot(""$G$"", (0,15), SE); dot(""$O$"", (0,0), NE); draw((0,0)--(0, 15)); draw((-7.5,15)--(0,0)); [/asy]530According to the Pythagorean Theorem and the Vertical Theorem, we can find out that $OG=\sqrt{\left(\frac{2\times9+16}{2}\right)^2 - \left(\frac{16}{2}\right)^2}=15$ . Therefore, the answer is $15\times16=\boxed{240}$",240531163,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_19,3,"A number is called flippy if its digits alternate between two distinct digits. For example, $2020$ and $37373$ are flippy, but $3883$ and $123123$ are not. How many five-digit flippy numbers are divisible by $15?$532$\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }5 \qquad \textbf{(D) }6 \qquad \textbf{(E) }8$","After finding out that the last digit must be $5$ , the number is of the form $5\square 5\square 5$ . If the unknown digit is $x$ , we can find that one of the solutions to $x$ is $0$ , since $5+5+5$ is equal to $15$ , which is divisible by $3$ . After trying every one digit number, you'll notice that $x$ must be a multiple of $3$ , meaning that $x=0$ $3$ $6$ , or $9$ $50505$ $53535$ $56565$ , and $59595$ are the $\boxed{4}$ solutions to this question.",4533164,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_19,4,"A number is called flippy if its digits alternate between two distinct digits. For example, $2020$ and $37373$ are flippy, but $3883$ and $123123$ are not. How many five-digit flippy numbers are divisible by $15?$534$\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }5 \qquad \textbf{(D) }6 \qquad \textbf{(E) }8$","assume the number is $ababa$ $10101a+1010b=0 (mod 15)\newline$ $6a+5b=0 (mod 15)\newline$ $a=0 (mod 5)\newline$ $5b=0 (mod 15)\newline$ $b=0 (mod 3)\newline$ Solutions: $(5,0),(5,3),(5,6),(5,9)\newline$ $\boxed{4}$",4535165,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_21,1,"A game board consists of $64$ squares that alternate in color between black and white. The figure below shows square $P$ in the bottom row and square $Q$ in the top row. A marker is placed at $P.$ A step consists of moving the marker onto one of the adjoining white squares in the row above. How many $7$ -step paths are there from $P$ to $Q?$ (The figure shows a sample path.)536[asy]//diagram by SirCalcsALot size(200); int[] x = {6, 5, 4, 5, 6, 5, 6}; int[] y = {1, 2, 3, 4, 5, 6, 7}; int N = 7; for (int i = 0; i < 8; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } for (int i = 0; i < N; ++i) { draw(circle((x[i],y[i])+(0.5,0.5),0.35)); } label(""$P$"", (5.5, 0.5)); label(""$Q$"", (6.5, 7.5)); [/asy]537$\textbf{(A) }28 \qquad \textbf{(B) }30 \qquad \textbf{(C) }32 \qquad \textbf{(D) }33 \qquad \textbf{(E) }35$","Notice that, in order to step onto any particular white square, the marker must have come from one of the $1$ or $2$ white squares immediately beneath it (since the marker can only move on white squares). This means that the number of ways to move from $P$ to that square is the sum of the numbers of ways to move from $P$ to each of the white squares immediately beneath it(also called the Water Fall Method). To solve the problem, we can accordingly construct the following diagram, where each number in a square is calculated as the sum of the numbers on the white squares immediately beneath that square (and thus will represent the number of ways to remove from $P$ to that square, as already stated).538[asy] int N = 7; for (int i = 0; i < 8; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } label(""$1$"", (5.5, .5)); label(""$1$"", (4.5, 1.5)); label(""$1$"", (6.5, 1.5)); label(""$1$"", (3.5, 2.5)); label(""$1$"", (7.5, 2.5)); label(""$2$"", (5.5, 2.5)); label(""$1$"", (2.5, 3.5)); label(""$3$"", (6.5, 3.5)); label(""$3$"", (4.5, 3.5)); label(""$4$"", (3.5, 4.5)); label(""$3$"", (7.5, 4.5)); label(""$6$"", (5.5, 4.5)); label(""$10$"", (4.5, 5.5)); label(""$9$"", (6.5, 5.5)); label(""$19$"", (5.5, 6.5)); label(""$9$"", (7.5, 6.5)); label(""$28$"", (6.5, 7.5)); [/asy]539The answer is therefore $\boxed{28}$",28540166,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_21,2,"A game board consists of $64$ squares that alternate in color between black and white. The figure below shows square $P$ in the bottom row and square $Q$ in the top row. A marker is placed at $P.$ A step consists of moving the marker onto one of the adjoining white squares in the row above. How many $7$ -step paths are there from $P$ to $Q?$ (The figure shows a sample path.)541[asy]//diagram by SirCalcsALot size(200); int[] x = {6, 5, 4, 5, 6, 5, 6}; int[] y = {1, 2, 3, 4, 5, 6, 7}; int N = 7; for (int i = 0; i < 8; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } for (int i = 0; i < N; ++i) { draw(circle((x[i],y[i])+(0.5,0.5),0.35)); } label(""$P$"", (5.5, 0.5)); label(""$Q$"", (6.5, 7.5)); [/asy]542$\textbf{(A) }28 \qquad \textbf{(B) }30 \qquad \textbf{(C) }32 \qquad \textbf{(D) }33 \qquad \textbf{(E) }35$","Suppose we ""extend"" the chessboard infinitely with $2$ additional columns to the right, as shown below. The red line shows the right-hand edge of the original board.543[asy] int N = 7; for (int i = 0; i < 10; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } draw((8,0) -- (8,8),red); label(""$P$"", (5.5,.5)); label(""$Q$"", (6.5,7.5)); label(""$X$"", (8.5,3.5)); label(""$Y$"", (8.5,5.5)); [/asy]544The total number of paths from $P$ to $Q$ , including invalid paths which cross over the red line, is then the number of paths which make $4$ steps up-and-right and $3$ steps up-and-left, which is $\binom{4+3}{3} = \binom{7}{3} = 35$ . We need to subtract the number of invalid paths, i.e. the number of paths that pass through $X$ or $Y$ . To get to $X$ , the marker has to make $3$ up-and-right steps, after which it can proceed to $Q$ with $3$ steps up-and-left and $1$ step up-and-right. Thus, the number of paths from $P$ to $Q$ that pass through $X$ is $1 \cdot \binom{3+1}{3} = 4$ . Similarly, the number of paths that pass through $Y$ is $\binom{4+1}{1}\cdot 1 = 5$ . However, we have now double-counted the invalid paths which pass through both $X$ and $Y$ ; from the diagram, it is clear that there are only $2$ of these (as the marker can get from $X$ to $Y$ by a step up-and-left and a step-up-and-right in either order). Hence the number of invalid paths is $4+5-2=7$ , and the number of valid paths from $P$ to $Q$ is $35-7 = \boxed{28}$",28545167,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_21,3,"A game board consists of $64$ squares that alternate in color between black and white. The figure below shows square $P$ in the bottom row and square $Q$ in the top row. A marker is placed at $P.$ A step consists of moving the marker onto one of the adjoining white squares in the row above. How many $7$ -step paths are there from $P$ to $Q?$ (The figure shows a sample path.)546[asy]//diagram by SirCalcsALot size(200); int[] x = {6, 5, 4, 5, 6, 5, 6}; int[] y = {1, 2, 3, 4, 5, 6, 7}; int N = 7; for (int i = 0; i < 8; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } for (int i = 0; i < N; ++i) { draw(circle((x[i],y[i])+(0.5,0.5),0.35)); } label(""$P$"", (5.5, 0.5)); label(""$Q$"", (6.5, 7.5)); [/asy]547$\textbf{(A) }28 \qquad \textbf{(B) }30 \qquad \textbf{(C) }32 \qquad \textbf{(D) }33 \qquad \textbf{(E) }35$","On any white square, we may choose to go left or right, as long as we do not cross over the border of the board. Call the moves $L$ and $R$ respectively. Every single legal path consists of $4$ $R's$ and $3$ $L's$ , so now all we have to find is the number of ways to order $4 R's$ and $3 L's$ in any way, which is ${7 \choose 3}=35$ . However, we originally promised that we will not go over the border, and now we have to subtract the paths that do go over the border. The paths that go over the border are any paths that start with RRR(1 path), RR(5 paths) and LRRR(1 path) so our final number of paths is $35-7=\boxed{28}.$ ~PEKKA",28548168,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_22,1,"When a positive integer $N$ is fed into a machine, the output is a number calculated according to the rule shown below.549[asy] size(300); defaultpen(linewidth(0.8)+fontsize(13)); real r = 0.05; draw((0.9,0)--(3.5,0),EndArrow(size=7)); filldraw((4,2.5)--(7,2.5)--(7,-2.5)--(4,-2.5)--cycle,gray(0.65)); fill(circle((5.5,1.25),0.8),white); fill(circle((5.5,1.25),0.5),gray(0.65)); fill((4.3,-r)--(6.7,-r)--(6.7,-1-r)--(4.3,-1-r)--cycle,white); fill((4.3,-1.25+r)--(6.7,-1.25+r)--(6.7,-2.25+r)--(4.3,-2.25+r)--cycle,white); fill((4.6,-0.25-r)--(6.4,-0.25-r)--(6.4,-0.75-r)--(4.6,-0.75-r)--cycle,gray(0.65)); fill((4.6,-1.5+r)--(6.4,-1.5+r)--(6.4,-2+r)--(4.6,-2+r)--cycle,gray(0.65)); label(""$N$"",(0.45,0)); draw((7.5,1.25)--(11.25,1.25),EndArrow(size=7)); draw((7.5,-1.25)--(11.25,-1.25),EndArrow(size=7)); label(""if $N$ is even"",(9.25,1.25),N); label(""if $N$ is odd"",(9.25,-1.25),N); label(""$\frac N2$"",(12,1.25)); label(""$3N+1$"",(12.6,-1.25)); [/asy] For example, starting with an input of $N=7,$ the machine will output $3 \cdot 7 +1 = 22.$ Then if the output is repeatedly inserted into the machine five more times, the final output is $26.$ \[7 \to 22 \to 11 \to 34 \to 17 \to 52 \to 26\] When the same $6$ -step process is applied to a different starting value of $N,$ the final output is $1.$ What is the sum of all such integers $N?$ \[N \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to 1\] $\textbf{(A) }73 \qquad \textbf{(B) }74 \qquad \textbf{(C) }75 \qquad \textbf{(D) }82 \qquad \textbf{(E) }83$","We start with final output of $1$ and work backward, taking cares to consider all possible inputs that could have resulted in any particular output. This produces following set of possibilities each stage: \[\{1\}\rightarrow\{2\}\rightarrow\{4\}\rightarrow\{1,8\}\rightarrow\{2,16\}\rightarrow\{4,5,32\}\rightarrow\{1,8,10,64\}\] where, for example, $2$ must come from $4$ (as there is no integer $n$ satisfying $3n+1=2$ ), but $16$ could come from $32$ or $5$ (as $\frac{32}{2} = 3 \cdot 5 + 1 = 16$ , and $32$ is even while $5$ is odd). By construction, last set in this sequence contains all the numbers which will lead to number $1$ to end of the $6$ -step process, and sum is $1+8+10+64=\boxed{83}$",83550169,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_22,2,"When a positive integer $N$ is fed into a machine, the output is a number calculated according to the rule shown below.551[asy] size(300); defaultpen(linewidth(0.8)+fontsize(13)); real r = 0.05; draw((0.9,0)--(3.5,0),EndArrow(size=7)); filldraw((4,2.5)--(7,2.5)--(7,-2.5)--(4,-2.5)--cycle,gray(0.65)); fill(circle((5.5,1.25),0.8),white); fill(circle((5.5,1.25),0.5),gray(0.65)); fill((4.3,-r)--(6.7,-r)--(6.7,-1-r)--(4.3,-1-r)--cycle,white); fill((4.3,-1.25+r)--(6.7,-1.25+r)--(6.7,-2.25+r)--(4.3,-2.25+r)--cycle,white); fill((4.6,-0.25-r)--(6.4,-0.25-r)--(6.4,-0.75-r)--(4.6,-0.75-r)--cycle,gray(0.65)); fill((4.6,-1.5+r)--(6.4,-1.5+r)--(6.4,-2+r)--(4.6,-2+r)--cycle,gray(0.65)); label(""$N$"",(0.45,0)); draw((7.5,1.25)--(11.25,1.25),EndArrow(size=7)); draw((7.5,-1.25)--(11.25,-1.25),EndArrow(size=7)); label(""if $N$ is even"",(9.25,1.25),N); label(""if $N$ is odd"",(9.25,-1.25),N); label(""$\frac N2$"",(12,1.25)); label(""$3N+1$"",(12.6,-1.25)); [/asy] For example, starting with an input of $N=7,$ the machine will output $3 \cdot 7 +1 = 22.$ Then if the output is repeatedly inserted into the machine five more times, the final output is $26.$ \[7 \to 22 \to 11 \to 34 \to 17 \to 52 \to 26\] When the same $6$ -step process is applied to a different starting value of $N,$ the final output is $1.$ What is the sum of all such integers $N?$ \[N \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to 1\] $\textbf{(A) }73 \qquad \textbf{(B) }74 \qquad \textbf{(C) }75 \qquad \textbf{(D) }82 \qquad \textbf{(E) }83$","As in Solution 1, we work backwards from $1$ , this time showing the possible cases in a tree diagram:552[asy] // Upper branches draw((-6, 1.5)--(-5, 1)--(-3, 1)--(-2,0)--(0, 0)); draw((-6, 0.5)--(-5, 1)); // Lower branches draw((-6, -1.5)--(-5, -1.5)--(-4, -1)--(-3, -1)--(-2, 0)); draw((-6, -0.5)--(-5, -0.5)--(-4, -1));  label(""$1$"", (0, 0),  UnFill(0.1mm)); label(""$2$"", (-1, 0), UnFill(0.1mm)); label(""$4$"", (-2, 0), UnFill(0.1mm));  // Upper branches label(""$1$"", (-3, 1), UnFill(0.1mm)); label(""$2$"", (-4, 1), UnFill(0.1mm)); label(""$4$"", (-5, 1), UnFill(0.1mm)); label(""$\textbf{8}$"", (-6, 1.5), UnFill(0.1mm)); label(""$\textbf{1}$"", (-6, 0.5), UnFill(0.1mm));  // Lower branches label(""$8$"", (-3, -1), UnFill(0.1mm)); label(""$16$"",(-4, -1), UnFill(0.1mm)); label(""$5$"", (-5, -0.5), UnFill(0.1mm)); label(""$32$"", (-5, -1.5), UnFill(0.1mm)); label(""$\textbf{10}$"", (-6, -0.5), UnFill(0.1mm)); label(""$\textbf{64}$"", (-6, -1.5), UnFill(0.1mm)); [/asy]553The possible numbers are those at the ""leaves"" of the tree (the ends of the various branches), which are $1$ $8$ $64$ , and $10$ . Thus the answer is $1+8+64+10=\boxed{83}$",83554170,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_22,3,"When a positive integer $N$ is fed into a machine, the output is a number calculated according to the rule shown below.555[asy] size(300); defaultpen(linewidth(0.8)+fontsize(13)); real r = 0.05; draw((0.9,0)--(3.5,0),EndArrow(size=7)); filldraw((4,2.5)--(7,2.5)--(7,-2.5)--(4,-2.5)--cycle,gray(0.65)); fill(circle((5.5,1.25),0.8),white); fill(circle((5.5,1.25),0.5),gray(0.65)); fill((4.3,-r)--(6.7,-r)--(6.7,-1-r)--(4.3,-1-r)--cycle,white); fill((4.3,-1.25+r)--(6.7,-1.25+r)--(6.7,-2.25+r)--(4.3,-2.25+r)--cycle,white); fill((4.6,-0.25-r)--(6.4,-0.25-r)--(6.4,-0.75-r)--(4.6,-0.75-r)--cycle,gray(0.65)); fill((4.6,-1.5+r)--(6.4,-1.5+r)--(6.4,-2+r)--(4.6,-2+r)--cycle,gray(0.65)); label(""$N$"",(0.45,0)); draw((7.5,1.25)--(11.25,1.25),EndArrow(size=7)); draw((7.5,-1.25)--(11.25,-1.25),EndArrow(size=7)); label(""if $N$ is even"",(9.25,1.25),N); label(""if $N$ is odd"",(9.25,-1.25),N); label(""$\frac N2$"",(12,1.25)); label(""$3N+1$"",(12.6,-1.25)); [/asy] For example, starting with an input of $N=7,$ the machine will output $3 \cdot 7 +1 = 22.$ Then if the output is repeatedly inserted into the machine five more times, the final output is $26.$ \[7 \to 22 \to 11 \to 34 \to 17 \to 52 \to 26\] When the same $6$ -step process is applied to a different starting value of $N,$ the final output is $1.$ What is the sum of all such integers $N?$ \[N \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to 1\] $\textbf{(A) }73 \qquad \textbf{(B) }74 \qquad \textbf{(C) }75 \qquad \textbf{(D) }82 \qquad \textbf{(E) }83$","We begin by finding the inverse of the function that the machine uses. Call the input $I$ and the output $O$ . If $I$ is even, $O=\frac{I}{2}$ , and if $I$ is odd, $O=3I+1$ . We can therefore see that $I=2O$ when $I$ is even and $I=\frac{O-1}{3}$ when $I$ is odd. Therefore, starting with $1$ , if $I$ is even, $I=2$ , and if $I$ is odd, $I=0$ , but the latter is not valid since $0$ is not actually odd. This means that the 2nd-to-last number in the sequence has to be $2$ . Now, substituting $2$ into the inverse formulae, if $I$ is even, $I=4$ (which is indeed even), and if $I$ is odd, $I=\frac{1}{3}$ , which is not an integer. This means the 3rd-to-last number in the sequence has to be $4$ . Substituting in $4$ , if $I$ is even, $I=8$ , but if $I$ is odd, $I=1$ . Both of these are valid solutions, so the 4th-to-last number can be either $1$ or $8$ . If it is $1$ , then by the argument we have just made, the 5th-to-last number has to be $2$ , the 6th-to-last number has to be $4$ , and the 7th-to-last number, which is the first number, must be either $1$ or $8$ . In this way, we have ultimately found two solutions: $N=1$ and $N=8$556On the other hand, if the 4th-to-last number is $8$ , substituting $8$ into the inverse formulae shows that the 5th-to-last number is either $16$ or $\frac{7}{3}$ , but the latter is not an integer. Substituting $16$ shows that if $I$ is even, $I=32$ , but if I is odd, $I=5$ , and both of these are valid. If the 6th-to-last number is $32$ , then the first number must be $64$ , since $\frac{31}{3}$ is not an integer; if the 6th-to-last number is $5,$ then the first number has to be $10$ , as $\frac{4}{3}$ is not an integer. This means that, in total, there are $4$ solutions for $N$ , specifically, $1$ $8$ $10$ , and $64$ , which sum to $\boxed{83}$",83557171,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_23,1,"Five different awards are to be given to three students. Each student will receive at least one award. In how many different ways can the awards be distributed?558$\textbf{(A) }120 \qquad \textbf{(B) }150 \qquad \textbf{(C) }180 \qquad \textbf{(D) }210 \qquad \textbf{(E) }240$","Firstly, observe that it is not possible for a single student to receive $4$ or $5$ awards because this would mean that one of the other students receives no awards. Thus, each student must receive either $1$ $2$ , or $3$ awards. If a student receives $3$ awards, then the other two students must each receive $1$ award; if a student receives $2$ awards, then another student must also receive $2$ awards and the remaining student must receive $1$ award. We consider each of these two cases in turn.559If a student receives three awards, there are $3$ ways to choose which student this is, and $\binom{5}{3}$ ways to give that student $3$ out of the $5$ awards. Next, there are $2$ students left and $2$ awards to give out, with each student getting one award. There are clearly just $2$ ways to distribute these two awards out, giving $3\cdot\binom{5}{3}\cdot 2=60$ ways to distribute the awards in this case.560In the other case, two students receive $2$ awards and one student recieves $1$ award . We know there are $3$ choices for which student gets $1$ award. There are $\binom{3}{1}$ ways to do this. Then, there are $\binom{5}{2}$ ways to give the first student his two awards, leaving $3$ awards yet to distribute. There are then $\binom{3}{2}$ ways to give the second student his $2$ awards. Finally, there is only $1$ student and $1$ award left, so there is only $1$ way to distribute this award. This results in $\binom{5}{2}\cdot\binom{3}{2}\cdot 1\cdot 3 =90$ ways to distribute the awards in this case. Adding the results of these two cases, we get $60+90=\boxed{150}$",150561172,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_23,2,"Five different awards are to be given to three students. Each student will receive at least one award. In how many different ways can the awards be distributed?562$\textbf{(A) }120 \qquad \textbf{(B) }150 \qquad \textbf{(C) }180 \qquad \textbf{(D) }210 \qquad \textbf{(E) }240$","Upon inspection (specified in the above solution), there are two cases of the distribution of awards to the students: one student gets 3 awards and the other each get 1 award or one student gets 1 award and the other two get 2 awards.563In the first case, there are $\binom{3}{1} = 3$ ways to choose the person who gets 3 awards. From here, there are $\binom{5}{3} = 10$ ways to choose the 3 awards from the 5 total awards. Now, one person has $2$ choices for awards and the other has $1$ choice for the award. Thus, the total number of ways to choose awards in this case is $3 \cdot 10 \cdot 2 \cdot 1 = 60$564In the other case, there are $\binom{3}{1} = 3$ ways to choose the person who gets 1 award, and $5$ choices for his/her award. Then, one person has $\binom{4}{2} = 6$ ways to have his/her awards and the other person has $\dbinom{2}{2} = 1$ ways to have his/her awards. This gives $3 \cdot 5 \cdot 6 \cdot 1  = 90$ ways for this case.565Adding these cases together, we get $60 + 90 = 150$ ways to distribute the awards, or choice $\boxed{150}$",150566173,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_23,3,"Five different awards are to be given to three students. Each student will receive at least one award. In how many different ways can the awards be distributed?567$\textbf{(A) }120 \qquad \textbf{(B) }150 \qquad \textbf{(C) }180 \qquad \textbf{(D) }210 \qquad \textbf{(E) }240$","Without the restriction that each student receives at least one award, we could take each of the awards and choose one of the $3$ students to give it to. This would be $3^5$ ways to distribute the awards in total. Now we need to subtract the cases where at least one student doesn't receive an award. If a student doesn't receive an award, there are $3$ choices for which student that is, so $2^5$ ways of choosing a student to receive each of the awards; in total, $3\cdot32=96$568However, if $2$ students both don't receive an award, then this case would be counted twice among the $96$ , so we need to add back in these cases. Said in other words, $2$ students not receiving an award is equivalent to $1$ student receiving $5$ awards, and there are $3$ choices for whom that student would be. To finish, the total number of ways to distribute the awards is $243 - 96+3$ , or $\boxed{150}$",150569174,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_25,1,"Rectangles $R_1$ and $R_2,$ and squares $S_1,\,S_2,\,$ and $S_3,$ shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of $S_2$ in units?570[asy] draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0)); draw((3,0)--(3,1)--(0,1)); draw((3,1)--(3,2)--(5,2)); draw((3,2)--(2,2)--(2,1)--(2,3)); label(""$R_1$"",(3/2,1/2)); label(""$S_3$"",(4,1)); label(""$S_2$"",(5/2,3/2)); label(""$S_1$"",(1,2)); label(""$R_2$"",(7/2,5/2)); [/asy]571$\textbf{(A) }651 \qquad \textbf{(B) }655 \qquad \textbf{(C) }656 \qquad \textbf{(D) }662 \qquad \textbf{(E) }666$","Let the side length of each square $S_k$ be $s_k$ . Then, from the diagram, we can line up the top horizontal lengths of $S_1$ $S_2$ , and $S_3$ to cover the top side of the large rectangle, so $s_{1}+s_{2}+s_{3}=3322$ . Similarly, the short side of $R_2$ will be $s_1-s_2$ , and lining this up with the left side of $S_3$ to cover the vertical side of the large rectangle gives $s_{1}-s_{2}+s_{3}=2020$ . We subtract the second equation from the first to obtain $2s_{2}=1302$ , and thus $s_{2}=\boxed{651}$",651572175,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_25,2,"Rectangles $R_1$ and $R_2,$ and squares $S_1,\,S_2,\,$ and $S_3,$ shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of $S_2$ in units?573[asy] draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0)); draw((3,0)--(3,1)--(0,1)); draw((3,1)--(3,2)--(5,2)); draw((3,2)--(2,2)--(2,1)--(2,3)); label(""$R_1$"",(3/2,1/2)); label(""$S_3$"",(4,1)); label(""$S_2$"",(5/2,3/2)); label(""$S_1$"",(1,2)); label(""$R_2$"",(7/2,5/2)); [/asy]574$\textbf{(A) }651 \qquad \textbf{(B) }655 \qquad \textbf{(C) }656 \qquad \textbf{(D) }662 \qquad \textbf{(E) }666$","Assuming that the problem is well-posed, it should be true in the particular case where $S_1 \cong S_3$ and $R_1 \cong R_2$ . Let the sum of the side lengths of $S_1$ and $S_3$ be $x$ , and let the length of square $S_2$ be $y$ . We then have the system \[\begin{dcases}x+y =3322 \\x-y=2020\end{dcases}\] which we solve to determine $y=\boxed{651}$",651575176,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_25,3,"Rectangles $R_1$ and $R_2,$ and squares $S_1,\,S_2,\,$ and $S_3,$ shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of $S_2$ in units?576[asy] draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0)); draw((3,0)--(3,1)--(0,1)); draw((3,1)--(3,2)--(5,2)); draw((3,2)--(2,2)--(2,1)--(2,3)); label(""$R_1$"",(3/2,1/2)); label(""$S_3$"",(4,1)); label(""$S_2$"",(5/2,3/2)); label(""$S_1$"",(1,2)); label(""$R_2$"",(7/2,5/2)); [/asy]577$\textbf{(A) }651 \qquad \textbf{(B) }655 \qquad \textbf{(C) }656 \qquad \textbf{(D) }662 \qquad \textbf{(E) }666$","Since, for each pair of rectangles, the side lengths have a sum of $3322$ or $2020$ and a difference of $S_2$ , the answer must be $\dfrac{3322 - 2020}{2} = \dfrac{1302}{2} = \boxed{651}$",651578177,https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_25,4,"Rectangles $R_1$ and $R_2,$ and squares $S_1,\,S_2,\,$ and $S_3,$ shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of $S_2$ in units?579[asy] draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0)); draw((3,0)--(3,1)--(0,1)); draw((3,1)--(3,2)--(5,2)); draw((3,2)--(2,2)--(2,1)--(2,3)); label(""$R_1$"",(3/2,1/2)); label(""$S_3$"",(4,1)); label(""$S_2$"",(5/2,3/2)); label(""$S_1$"",(1,2)); label(""$R_2$"",(7/2,5/2)); [/asy]580$\textbf{(A) }651 \qquad \textbf{(B) }655 \qquad \textbf{(C) }656 \qquad \textbf{(D) }662 \qquad \textbf{(E) }666$","Let the side length of $S_2$ be s, and the shorter side length of $R_1$ and $R_2$ be $r$ . We have581[asy] draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0)); draw((3,0)--(3,1)--(0,1)); draw((3,1)--(3,2)--(5,2)); draw((3,2)--(2,2)--(2,1)--(2,3)); label(""$R_1$"",(3/2,1/2)); label(""$S_3$"",(4,1)); label(""$S_2$"",(5/2,3/2)); label(""$S_1$"",(1,2)); label(""$R_2$"",(7/2,5/2)); label(""$r$"",(5.2,5/2)); label(""$r$"",(3.2,1/2)); label(""$s$"",(3.2,3/2)); [/asy]582From this diagram, it is evident that $r+s+r=2020$ . Also, the side length of $S_1$ and $S_3$ is $r+s$ . Then, $r+s+s+r+s=3322$ . Now, we have 2 systems of equations.583\begin{align*}r+s+r &= 2020 \\ r+s+r+s+s &= 3322 \\ \end{align*}584We can see an $r+s+r$ in the 2nd equation, so substituting that in gives us $2020+2s=3322 \Rightarrow 2s= 1302 \Rightarrow s=\boxed{651}$",651585178,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_1,1,"Ike and Mike go into a sandwich shop with a total of $$30.00$ to spend. Sandwiches cost $$4.50$ each and soft drinks cost $$1.00$ each. Ike and Mike plan to buy as many sandwiches as they can,586and use any remaining money to buy soft drinks. Counting both sandwiches and soft drinks, how587many items will they buy?588$\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad\textbf{(E) }10$","We know that the sandwiches cost $4.50$ dollars. Guessing will bring us to multiplying $4.50$ by 6, which gives us $27.00$ . Since they can spend $30.00$ they have $3$ dollars left. Since sodas cost $1.00$ dollar each, they can buy 3 sodas, which makes them spend $30.00$ Since they bought 6 sandwiches and 3 sodas, they bought a total of $9$ items. Therefore, the answer is $\boxed{9}$",9589179,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_1,2,"Ike and Mike go into a sandwich shop with a total of $$30.00$ to spend. Sandwiches cost $$4.50$ each and soft drinks cost $$1.00$ each. Ike and Mike plan to buy as many sandwiches as they can,590and use any remaining money to buy soft drinks. Counting both sandwiches and soft drinks, how591many items will they buy?592$\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad\textbf{(E) }10$","Let $s$ be the number of sandwiches and $d$ be the number of sodas. We have to satisfy the equation of \[4.50s+d=30\] In the question, it states that Ike and Mike buys as many sandwiches as possible. 593So, we drop the number of sodas for a while.594We have: \begin{align*} 4.50s&=30 \\ s&=\frac{30}{4.5} \\ s&=6R3 \end{align*} We don't want a remainder so the maximum number of sandwiches is $6$ .595The total money spent is $6\cdot 4.50=27$ .596The number of dollar left to spent on sodas is $30-27=3$ dollars. $3$ dollars can buy $3$ sodas leading us to a total of $6+3=9$ items. 597Hence, the answer is $\boxed{9}$",9598180,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_2,1,"Three identical rectangles are put together to form rectangle $ABCD$ , as shown in the figure below. Given that the length of the shorter side of each of the smaller rectangles is  5 feet, what is the area in square feet of rectangle $ABCD$599[asy] draw((0,0)--(3,0)); draw((0,0)--(0,2)); draw((0,2)--(3,2)); draw((3,2)--(3,0)); dot((0,0)); dot((0,2)); dot((3,0)); dot((3,2)); draw((2,0)--(2,2)); draw((0,1)--(2,1)); label(""A"",(0,0),S); label(""B"",(3,0),S); label(""C"",(3,2),N); label(""D"",(0,2),N); [/asy]600$\textbf{(A) }45\qquad\textbf{(B) }75\qquad\textbf{(C) }100\qquad\textbf{(D) }125\qquad\textbf{(E) }150$","We can see that there are $2$ rectangles lying on top of the other and that is the same as the length of one rectangle. Now we know that the shorter side is $5$ , so the bigger side is $10$ , if we do $5 \cdot 2 = 10$ . Now we get the sides of the big rectangle being $15$ and $10$ , so the area is $\boxed{150}$ . ~avamarora",150601181,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_4,1,"Quadrilateral $ABCD$ is a rhombus with perimeter $52$ meters. The length of diagonal $\overline{AC}$ is $24$ meters. What is the area in square meters of rhombus $ABCD$602[asy] draw((-13,0)--(0,5)); draw((0,5)--(13,0)); draw((13,0)--(0,-5)); draw((0,-5)--(-13,0)); dot((-13,0)); dot((0,5)); dot((13,0)); dot((0,-5)); label(""A"",(-13,0),W); label(""B"",(0,5),N); label(""C"",(13,0),E); label(""D"",(0,-5),S); [/asy]603$\textbf{(A) }60\qquad\textbf{(B) }90\qquad\textbf{(C) }105\qquad\textbf{(D) }120\qquad\textbf{(E) }144$","[asy] draw((-12,0)--(0,5)); draw((0,5)--(12,0)); draw((12,0)--(0,-5)); draw((0,-5)--(-12,0)); draw((0,0)--(12,0)); draw((0,0)--(0,5)); draw((0,0)--(-12,0)); draw((0,0)--(0,-5)); dot((-12,0)); dot((0,5)); dot((12,0)); dot((0,-5)); label(""A"",(-12,0),W); label(""B"",(0,5),N); label(""C"",(12,0),E); label(""D"",(0,-5),S); label(""E"",(0,0),SW); [/asy]604A rhombus has sides of equal length. Because the perimeter of the rhombus is $52$ , each side is $\frac{52}{4}=13$ . In a rhombus, diagonals are perpendicular and bisect each other, which means $\overline{AE}$ $12$ $\overline{EC}$605Consider one of the right triangles:606[asy] draw((-12,0)--(0,5)); draw((0,0)--(-12,0)); draw((0,0)--(0,5)); dot((-12,0)); dot((0,5)); label(""A"",(-12,0),W); label(""B"",(0,5),N); label(""E"",(0,0),SE); [/asy]607$\overline{AB}$ $13$ , and $\overline{AE}$ $12$ . Using the Pythagorean theorem, we find that $\overline{BE}$ $5$ .608You know the Pythagorean triple, (5, 12, 13).609Thus the values of the two diagonals are $\overline{AC}$ $24$ and $\overline{BD}$ $10$ .610The area of a rhombus is = $\frac{d_1\cdot{d_2}}{2}$ $\frac{24\cdot{10}}{2}$ $120$611$\boxed{120}$",120612182,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_4,2,"Quadrilateral $ABCD$ is a rhombus with perimeter $52$ meters. The length of diagonal $\overline{AC}$ is $24$ meters. What is the area in square meters of rhombus $ABCD$613[asy] draw((-13,0)--(0,5)); draw((0,5)--(13,0)); draw((13,0)--(0,-5)); draw((0,-5)--(-13,0)); dot((-13,0)); dot((0,5)); dot((13,0)); dot((0,-5)); label(""A"",(-13,0),W); label(""B"",(0,5),N); label(""C"",(13,0),E); label(""D"",(0,-5),S); [/asy]614$\textbf{(A) }60\qquad\textbf{(B) }90\qquad\textbf{(C) }105\qquad\textbf{(D) }120\qquad\textbf{(E) }144$","Right off the bat, we can see that the perimeter of the figure is 52. Dividing this by four, we can get that each side is equal to 13. By drawing a line perpendicular to the one given, we can split the figure into four right triangles. 12 (24/2) is equal to the height of one small right triangle, and 13 is the slanted side. Using the Pythagorean theorem we can find that 169 (13 squared) - 144 (12 squared) = 25 (five squared). With this, we can determine that each small right triangle equals 30. Multiplying that by four we can get $\boxed{120}$",120615183,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_7,1,"Shauna takes five tests, each worth a maximum of $100$ points. Her scores on the first three tests are $76$ $94$ , and $87$ . In order to average $81$ for all five tests, what is the lowest score she could earn on one of the other two tests?616$\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }66\qquad\textbf{(D) }70\qquad\textbf{(E) }74$","We should notice that we can turn the information we are given into a linear equation and just solve for our set variables. I'll use the variables $x$ and $y$ for the scores on the last two tests. \[\frac{76+94+87+x+y}{5} = 81,\] \[\frac{257+x+y}{5} = 81.\] We can now cross multiply to get rid of the denominator. \[257+x+y = 405,\] \[x+y = 148.\] Now that we have this equation, we will assign $y$ as the lowest score of the two other tests, and so: \[x = 100,\] \[y=48.\] Now we know that the lowest score on the two other tests is $\boxed{48}$",48617184,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_7,2,"Shauna takes five tests, each worth a maximum of $100$ points. Her scores on the first three tests are $76$ $94$ , and $87$ . In order to average $81$ for all five tests, what is the lowest score she could earn on one of the other two tests?618$\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }66\qquad\textbf{(D) }70\qquad\textbf{(E) }74$","Right now, she scored $76, 94,$ and $87$ points, for a total of $257$ points. She wants her average to be $81$ for her $5$ tests, so she needs to score $405$ points in total. This means she needs to score a total of $405-257=  148$ points in her next $2$ tests. Since the maximum score she can get on one of her $2$ tests is $100$ , the least possible score she can get is $\boxed{48}$",48619185,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_7,3,"Shauna takes five tests, each worth a maximum of $100$ points. Her scores on the first three tests are $76$ $94$ , and $87$ . In order to average $81$ for all five tests, what is the lowest score she could earn on one of the other two tests?620$\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }66\qquad\textbf{(D) }70\qquad\textbf{(E) }74$","We can compare each of the scores with the average of $81$ $76$ $\rightarrow$ $-5$ $94$ $\rightarrow$ $+13$ $87$ $\rightarrow$ $+6$ $100$ $\rightarrow$ $+19$621So the last one has to be $-33$ (since all the differences have to sum to $0$ ), which corresponds to $81-33 = \boxed{48}$",48622186,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_8,1,"Gilda has a bag of marbles. She gives $20\%$ of them to her friend Pedro. Then Gilda gives $10\%$ of what is left to another friend, Ebony. Finally, Gilda gives $25\%$ of what is now left in the bag to her brother Jimmy. What percentage of her original bag of marbles does Gilda have left for herself?623$\textbf{(A) }20\qquad\textbf{(B) }33\frac{1}{3}\qquad\textbf{(C) }38\qquad\textbf{(D) }45\qquad\textbf{(E) }54$","After Gilda gives $20$ % of the marbles to Pedro, she has $80$ % of the marbles left. If she then gives $10$ % of what's left to Ebony, she has $(0.8*0.9)$ $72$ % of what she had at the beginning. Finally, she gives $25$ % of what's left to her brother, so she has $(0.75*0.72)$ $\boxed{54}$ of what she had in the beginning left.",54624187,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_8,2,"Gilda has a bag of marbles. She gives $20\%$ of them to her friend Pedro. Then Gilda gives $10\%$ of what is left to another friend, Ebony. Finally, Gilda gives $25\%$ of what is now left in the bag to her brother Jimmy. What percentage of her original bag of marbles does Gilda have left for herself?625$\textbf{(A) }20\qquad\textbf{(B) }33\frac{1}{3}\qquad\textbf{(C) }38\qquad\textbf{(D) }45\qquad\textbf{(E) }54$","Suppose Gilda has 100 marbles.626Then, she gives Pedro 20% of 100 = 20, she remains with 80 marbles.627Out of 80 marbles, she gives 10% of 80 = 8 to Ebony.628Thus, she remains with 72 marbles.629Then, she gives 25% of 72 = 18 to Jimmy, finally leaving her with 54.630And, $\frac{54}{100}$ =54%= $\boxed{54}$",54631188,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_8,3,"Gilda has a bag of marbles. She gives $20\%$ of them to her friend Pedro. Then Gilda gives $10\%$ of what is left to another friend, Ebony. Finally, Gilda gives $25\%$ of what is now left in the bag to her brother Jimmy. What percentage of her original bag of marbles does Gilda have left for herself?632$\textbf{(A) }20\qquad\textbf{(B) }33\frac{1}{3}\qquad\textbf{(C) }38\qquad\textbf{(D) }45\qquad\textbf{(E) }54$","(Only if you have lots of time do it this way)633Since she gave away 20% and 10% of what is left and then another 25% of what is actually left, we can do 20+10+25 or 55%. But it is actually going to be a bit more than 55% because 10% of what is left is not 10% of the total amount. So, the only option that is greater than 100% - 55% is $\boxed{54}$",54634189,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_11,1,"The eighth grade class at Lincoln Middle School has $93$ students. Each student takes a math class or a foreign language class or both. There are $70$ eighth graders taking a math class, and there are $54$ eighth graders taking a foreign language class. How many eighth graders take only a math class and not a foreign language class?635$\textbf{(A) }16\qquad\textbf{(B) }53\qquad\textbf{(C) }31\qquad\textbf{(D) }39\qquad\textbf{(E) }70$","Let $x$ be the number of students taking both a math and a foreign language class.636By P-I-E, we get $70 + 54 - x$ $93$637Solving gives us $x = 31$638But we want the number of students taking only a math class,639which is $70 - 31 = 39$640$\boxed{39}$",39641190,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_11,2,"The eighth grade class at Lincoln Middle School has $93$ students. Each student takes a math class or a foreign language class or both. There are $70$ eighth graders taking a math class, and there are $54$ eighth graders taking a foreign language class. How many eighth graders take only a math class and not a foreign language class?642$\textbf{(A) }16\qquad\textbf{(B) }53\qquad\textbf{(C) }31\qquad\textbf{(D) }39\qquad\textbf{(E) }70$","We have $70 + 54 = 124$ people taking classes. However, we over-counted the number of people who take both classes. If we subtract the original amount of people who take classes we get that $31$ people took the two classes. To find the amount of people who took only math class, we subtract the people who didn't take only one math class, so we get $70 - 31 = \boxed{39}$",39643191,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_11,3,"The eighth grade class at Lincoln Middle School has $93$ students. Each student takes a math class or a foreign language class or both. There are $70$ eighth graders taking a math class, and there are $54$ eighth graders taking a foreign language class. How many eighth graders take only a math class and not a foreign language class?644$\textbf{(A) }16\qquad\textbf{(B) }53\qquad\textbf{(C) }31\qquad\textbf{(D) }39\qquad\textbf{(E) }70$","[asy] draw(circle((-0.5,0),1)); draw(circle((0.5,0),1)); label(""$\huge{x}$"", (0, 0)); label(""$70-x$"", (-1, 0)); label(""$54-x$"", (1, 0)); [/asy]645We know that the sum of all three areas is $93$ So, we have: \[93 = 70-x+x+54-x\] \[93 = 70+54-x\] \[93 = 124 - x\] \[-31=-x\] \[x=31\]646We are looking for the number of students in only math. This is $70-x$ . Substituting $x$ with $31$ , our answer is $\boxed{39}$",39647192,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_13,1,"palindrome is a number that has the same value when read from left to right or from right to left. (For example, 12321 is a palindrome.) Let $N$ be the least three-digit integer which is not a palindrome but which is the sum of three distinct two-digit palindromes. What is the sum of the digits of $N$648$\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad\textbf{(E) }6$","Note that the only positive 2-digit palindromes are multiples of 11, namely $11, 22, \ldots, 99$ . Since $N$ is the sum of 2-digit palindromes, $N$ is necessarily a multiple of 11. The smallest 3-digit multiple of 11 which is not a palindrome is 110, so $N=110$ is a candidate solution. We must check that 110 can be written as the sum of three distinct 2-digit palindromes; this suffices as $110=77+22+11$ . Then, $N = 110$ , and the sum of the digits of $N$ is $1+1+0 = \boxed{2}$",2649193,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_13,2,"palindrome is a number that has the same value when read from left to right or from right to left. (For example, 12321 is a palindrome.) Let $N$ be the least three-digit integer which is not a palindrome but which is the sum of three distinct two-digit palindromes. What is the sum of the digits of $N$650$\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad\textbf{(E) }6$","We already know that two-digit palindromes can only be two-digit multiples of 11; which are: $11, 22, 33, 44, 55, 66, 77, 88,$ and $99$ . Since this is clear, we will need to find out the least multiple of 11 that is not a palindrome. Then, we start counting. $110 \ldots$ Aha! This multiple of 11, 110, not only isn’t a palindrome, but it also is the sum of three distinct two-digit palindromes, for example: 11 + 22 + 77, 22 + 33 + 55, and 44 + 11 + 55! The sum of $N$ ’s digits is $1+1+0 = \boxed{2}$",2651194,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_13,3,"palindrome is a number that has the same value when read from left to right or from right to left. (For example, 12321 is a palindrome.) Let $N$ be the least three-digit integer which is not a palindrome but which is the sum of three distinct two-digit palindromes. What is the sum of the digits of $N$652$\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad\textbf{(E) }6$","As stated above, two-digit palindromes can only be two-digit multiples of 11. We can see that if we add anything that are multiples of 11 together, we will again get a multiple of 11. For instance, $11+22=33$ . Since we know this fact and we are finding the smallest value possible, we can start with the first three-digit multiple of 11 which is $110$ . Since this is not a palindrome and can be the sum of 3 two-digit palindromes (see above solutions for more details), $110$ fits the bill. We can see that the sum of $110$ 's digits is $1+1+0 = \boxed{2}$",2653195,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_16,1,"Qiang drives $15$ miles at an average speed of $30$ miles per hour. How many additional miles will he have to drive at $55$ miles per hour to average $50$ miles per hour for the entire trip?654$\textbf{(A) }45\qquad\textbf{(B) }62\qquad\textbf{(C) }90\qquad\textbf{(D) }110\qquad\textbf{(E) }135$","The only option that is easily divisible by $55$ is $110$ , which gives 2 hours of travel. And, the formula is $\frac{15}{30} + \frac{110}{55} = \frac{5}{2}$655And, $\text{Average Speed}$ $\frac{\text{Total Distance}}{\text{Total Time}}$656Thus, $\frac{125}{50} = \frac{5}{2}$657Both are equal and thus our answer is $\boxed{110}.$",110658196,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_16,2,"Qiang drives $15$ miles at an average speed of $30$ miles per hour. How many additional miles will he have to drive at $55$ miles per hour to average $50$ miles per hour for the entire trip?659$\textbf{(A) }45\qquad\textbf{(B) }62\qquad\textbf{(C) }90\qquad\textbf{(D) }110\qquad\textbf{(E) }135$","To calculate the average speed, simply evaluate the total distance over the total time. Let the number of additional miles he has to drive be $x.$ Therefore, the total distance is $15+x$ and the total time (in hours) is \[\frac{15}{30}+\frac{x}{55}=\frac{1}{2}+\frac{x}{55}.\] We can set up the following equation: \[\frac{15+x}{\frac{1}{2}+\frac{x}{55}}=50.\] Simplifying the equation, we get \[15+x=25+\frac{10x}{11}.\] Solving the equation yields $x=110,$ so our answer is $\boxed{110}$",110660197,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_16,3,"Qiang drives $15$ miles at an average speed of $30$ miles per hour. How many additional miles will he have to drive at $55$ miles per hour to average $50$ miles per hour for the entire trip?661$\textbf{(A) }45\qquad\textbf{(B) }62\qquad\textbf{(C) }90\qquad\textbf{(D) }110\qquad\textbf{(E) }135$","If he travels $15$ miles at a speed of $30$ miles per hour, he travels for 30 min. Average rate is total distance over total time so $(15+d)/(0.5 + t) = 50$ , where d is the distance left to travel and t is the time to travel that distance. Solve for $d$ to get $d = 10+50t$ . You also know that he has to travel $55$ miles per hour for some time, so $d=55t$ . Plug that in for d to get $55t = 10+50t$ and $t=2$ and since $d=55t$ $d = 2\cdot55 =110$ , the answer is $\boxed{110}$",110662198,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_16,4,"Qiang drives $15$ miles at an average speed of $30$ miles per hour. How many additional miles will he have to drive at $55$ miles per hour to average $50$ miles per hour for the entire trip?663$\textbf{(A) }45\qquad\textbf{(B) }62\qquad\textbf{(C) }90\qquad\textbf{(D) }110\qquad\textbf{(E) }135$","Let $h$ be the amount of hours Qiang drives after his first 15 miles. Average speed, which we know is $50$ mph, means total distance over total time. For 15 miles at 30 mph, the time taken is $\frac{1}{2}$ hour, so the total time for this trip would be $\frac{1}{2} + h$ hours. For the total distance, 15 miles are traveled in the first part and $55h$ miles in the second. This gives the following equation:664\[\dfrac{15+55h}{\frac{1}{2}+h} = 50.\]665Cross multiplying, we get that $15 + 55h = 50h + 25$ , and simple algebra gives $h=2$ . In 2 hours traveling at 55 mph, the distance traveled is $\frac{2 \hspace{0.05 in} \text{hours}}{1} \cdot \frac{55 \hspace{0.05 in} \text{miles}}{1 \hspace{0.05 in} \text{hour}} = 2 \cdot 55 \hspace{0.05 in} \text{miles} = 110 \hspace{0.05 in} \text{miles}$ , which is choice $\boxed{110}$",110666199,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_19,1,"In a tournament there are six teams that play each other twice. A team earns $3$ points for a win, $1$ point for a draw, and $0$ points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for each of the top three teams?667$\textbf{(A) }22\qquad\textbf{(B) }23\qquad\textbf{(C) }24\qquad\textbf{(D) }26\qquad\textbf{(E) }30$","This isn't finished668to another. This gives equality, as each team wins once and loses once as well. For a win, we have $3$ points, so a team gets $3\times2=6$ points if they each win a game and lose a game. This case brings a total of $18+6=24$ points.669Therefore, we use Case 2  since it brings the greater amount of points, or $\boxed{24}$",24670200,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_19,2,"In a tournament there are six teams that play each other twice. A team earns $3$ points for a win, $1$ point for a draw, and $0$ points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for each of the top three teams?671$\textbf{(A) }22\qquad\textbf{(B) }23\qquad\textbf{(C) }24\qquad\textbf{(D) }26\qquad\textbf{(E) }30$","We can name the top three teams as $A$ $B$ , and $C$ . We can see that (respective scores of) $A=B=C$ because these teams have the same points. If we look at the matches that involve the top three teams, we see that there are some duplicates: $AB$ $BC$ , and $AC$ come twice. In order to even out the scores and get the maximum score, we can say that in match $AB$ $A$ and $B$ each win once out of the two games that they play. We can say the same thing for $AC$ and $BC$ . This tells us that each team $A$ $B$ , and $C$ win and lose twice. This gives each team a total of $3 + 3 + 0 + 0 = 6$ points. Now, we need to include the other three teams. We can label these teams as $D$ $E$ , and $F$ . We can write down every match that $A, B,$ or $C$ plays in that we haven't counted yet: $AD$ $AD$ $AE$ $AE$ $AF$ $AF$ $BD$ $BD$ $BE$ $BE$ $BF$ $BF$ $CD$ $CD$ $CE$ $CE$ $CF$ , and $CF$ . We can say $A$ $B$ , and $C$ win each of these in order to obtain the maximum score that $A$ $B$ , and $C$ can have. If $A$ $B$ , and $C$ win all six of their matches, $A$ $B$ , and $C$ will have a score of $18$ $18 + 6$ results in a maximum score of $\boxed{24}$",24672201,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_19,3,"In a tournament there are six teams that play each other twice. A team earns $3$ points for a win, $1$ point for a draw, and $0$ points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for each of the top three teams?673$\textbf{(A) }22\qquad\textbf{(B) }23\qquad\textbf{(C) }24\qquad\textbf{(D) }26\qquad\textbf{(E) }30$","To start, we calculate how many games each team plays. Each team can play against $5$ people twice, so there are $10$ games that each team plays. So the answer is $10\cdot 3$ which is $30!$ But wait... if we want $3$ teams to have the same amount of points, there can't possibly be a player who wins all their games. Let the top three teams be $A,B$ , and $C.$ $A$ plays $B$ and $C$ twice so in order to maximize the games being played, we can split it $50-50$ between the $4$ games $A$ plays against $B$ or $C$ . We find that we just subtract $2$ games or $6$ points. Therefore the answer is $30-6$ $24$ or $\boxed{24}$",24674202,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_20,1,"How many different real numbers $x$ satisfy the equation \[(x^{2}-5)^{2}=16?\]675$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }8$","We have that $(x^2-5)^2 = 16$ if and only if $x^2-5 = \pm 4$ . If $x^2-5 = 4$ , then $x^2 = 9 \implies x = \pm 3$ , giving 2 solutions. If $x^2-5 = -4$ , then $x^2 = 1 \implies x = \pm 1$ , giving 2 more solutions. All four of these solutions work, so the answer is $\boxed{4}$ . Further, the equation is a quartic in $x$ , so by the Fundamental Theorem of Algebra , there can be at most four real solutions.",4676203,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_20,2,"How many different real numbers $x$ satisfy the equation \[(x^{2}-5)^{2}=16?\]677$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }8$","We can expand $(x^2-5)^2$ to get $x^4-10x^2+25$ , so now our equation is $x^4-10x^2+25=16$ . Subtracting $16$ from both sides gives us $x^4-10x^2+9=0$ . Now, we can factor the left hand side to get $(x^2-9)(x^2-1)=0$ . If $x^2-9$ and/or $x^2-1$ equals $0$ , then the whole left side will equal $0$ . Since the solutions can be both positive and negative, we have $4$ solutions: $-3,3,-1,1$ (we can find these solutions by setting $x^2-9$ and $x^2-1$ equal to $0$ and solving for $x$ ). So, the answer is $\boxed{4}$",4678204,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_20,3,"How many different real numbers $x$ satisfy the equation \[(x^{2}-5)^{2}=16?\]679$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }8$","Subtract 16 from both sides and factor using difference of squares:680\[(x^2 - 5)^2 = 16\] \[(x^2 - 5)^2 - 16 =0\] \[(x^2 - 5)^2 - 4^2 = 0\] \[[(x^2 - 5)-4][(x^2 - 5) + 4] = 0\] \[(x^2 - 9)(x^2 - 1) =0\] \[(x+3)(x-3)(x+1)(x-1) = 0\]681Quite obviously, this equation has $\boxed{4}$ solutions.",4682205,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_21,2,"What is the area of the triangle formed by the lines $y=5$ $y=1+x$ , and $y=1-x$683$\textbf{(A) }4\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }12\qquad\textbf{(E) }16$","Graphing the lines, using the intersection points we found in Solution 1, we can see that the height of the triangle is 4, and the base is 8. Using the formula for the area of a triangle, we get $\frac{4\cdot8}{2}$ which is equal to $\boxed{16}$",16684206,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_21,3,"What is the area of the triangle formed by the lines $y=5$ $y=1+x$ , and $y=1-x$685$\textbf{(A) }4\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }12\qquad\textbf{(E) }16$","$y = x + 1$ and $y = -x + 1$ have $y$ -intercepts at $(0, 1)$ and slopes of $1$ and $-1$ , respectively. Since the product of these slopes is $-1$ , the two lines are perpendicular. From $y = 5$ , we see that $(-4, 5)$ and $(4, 5)$ are the other two intersection points, and they are $8$ units apart. By symmetry, this triangle is a $45-45-90$ triangle, so the legs are $4\sqrt{2}$ each and the area is $\frac{(4\sqrt{2})^2}{2} = \boxed{16}$",16686207,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_22,1,"A store increased the original price of a shirt by a certain percent and then lowered the new price by the same amount. Given that the resulting price was $84\%$ of the original price, by what percent was the price increased and decreased $?$687$\textbf{(A) }16\qquad\textbf{(B) }20\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }40$","Suppose the fraction of discount is $x$ . That means $(1-x)(1+x)=0.84$ ; so, $1-x^{2}=0.84$ , and $(x^{2})=0.16$ , obtaining $x=0.4$ . Therefore, the price was increased and decreased by $40$ %, or $\boxed{40}$",40688208,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_22,2,"A store increased the original price of a shirt by a certain percent and then lowered the new price by the same amount. Given that the resulting price was $84\%$ of the original price, by what percent was the price increased and decreased $?$689$\textbf{(A) }16\qquad\textbf{(B) }20\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }40$","After the first increase by $p$ percent, the shirt price became $(1+p)$ times greater than the original. Upon the decrease in p percent on this price, the shirt price became $(1-p)$ times less than $(1+p)$ , or $(1-p)(1+p)$ . We know that this price is $84$ percent of the original, so $(1-p)(1+p) = 0.84$690From here, we can list the factors of $0.84$ and see which are equidistant from $1$ . We see that $0.6$ and $1.4$ are both $0.4$ from $1$ , so $p = 0.4 = 40 \%$ , or choice $\boxed{40}$",40691209,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_22,3,"A store increased the original price of a shirt by a certain percent and then lowered the new price by the same amount. Given that the resulting price was $84\%$ of the original price, by what percent was the price increased and decreased $?$692$\textbf{(A) }16\qquad\textbf{(B) }20\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }40$","We can try out every option and see which one works. By this method, we get $\boxed{40}$",40693210,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_22,5,"A store increased the original price of a shirt by a certain percent and then lowered the new price by the same amount. Given that the resulting price was $84\%$ of the original price, by what percent was the price increased and decreased $?$694$\textbf{(A) }16\qquad\textbf{(B) }20\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }40$","Let our original cost be $$ 100$ , so we are looking for a whole number of $$ 84$ . Also, we can see that (A), (C), and (D) give us answers with decimals while we know that (B) and (E) give us whole numbers. Therefore, we only need to try these two: (B) $$100$ increased by 20% = $$120$ , and $$120$ decreased by 20% = $$96$ , a whole number, and (E) $$100$ increased by 40% = $$140$ , and $$140$ decreased by 40% = $$84$ , a whole number.695Thus, $40$ % or $\boxed{40}$ is the answer.",40696211,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_23,1,"After Euclid High School's last basketball game, it was determined that $\frac{1}{4}$ of the team's points were scored by Alexa and $\frac{2}{7}$ were scored by Brittany. Chelsea scored $15$ points. None of the other $7$ team members scored more than $2$ points. What was the total number of points scored by the other $7$ team members?697$\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14$","Given the information above, we start with the equation $\frac{t}{4}+\frac{2t}{7} + 15 + x = t$ ,where $t$ is the total number of points scored and $x\le 14$ is the number of points scored by the remaining 7 team members, we can simplify to obtain the Diophantine equation $x+15 = \frac{13}{28}t$ , or $28x+28\cdot 15=13t$ . Since $t$ is necessarily divisible by 28, let $t=28u$ where $u \ge 0$ and divide by 28 to obtain $x + 15 = 13u$ . Then, it is easy to see $u=2$ $t=56$ ) is the only candidate remaining, giving $x=\boxed{11}$",11698212,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_23,3,"After Euclid High School's last basketball game, it was determined that $\frac{1}{4}$ of the team's points were scored by Alexa and $\frac{2}{7}$ were scored by Brittany. Chelsea scored $15$ points. None of the other $7$ team members scored more than $2$ points. What was the total number of points scored by the other $7$ team members?699$\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14$","Adding together Alexa's and Brittany's fractions, we get $\frac{15}{28}$ as the fraction of the total number of points they scored together. However, this is just a ratio, so we can introduce a variable: $\frac{15x}{28x}$ where $x$ is the common ratio. Let $y$ and $z$ and $w$ be the number of people who scored 1, 2, and 0 points, respectively. Writing an equation, we have $\frac{13x}{28x} = 15 + y + 2z + 0w.$ We want all of our variables to be integers. Thus, we want $15 + y + 2z = 0 \pmod {13}.$ Simplifying, $y+2z = 11 \pmod {13}.$ The only possible value, as this integer sum has to be less than $7 \cdot 2 + 1 = 15,$ must be 11. Therefore, $y+2z = 11,$ and the answer is $\boxed{11}$",11700213,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_23,4,"After Euclid High School's last basketball game, it was determined that $\frac{1}{4}$ of the team's points were scored by Alexa and $\frac{2}{7}$ were scored by Brittany. Chelsea scored $15$ points. None of the other $7$ team members scored more than $2$ points. What was the total number of points scored by the other $7$ team members?701$\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14$","We can rewrite the question as an algebraic equation: $\frac{1}{4} x + \frac{2}{7} x + 15 + y$ , where $x$ represents the total amount of points and $y$ the amount of points the $7$ other players scored. From there, we add the two fractions to get $\frac{15}{28} x + 15 = x$ . Subtracting $\frac{15}{28} x$ from both sides, we get $\frac{13}{28} x = y + 15$ . We multiply each side by $28$ to get rid of the denominator, in which we get $13x = 420 + 28y$ . Now let’s think of this logically. This equation is telling us that if you add $420$ and $28$ times the amount of points scored by the extra $7$ players, you get $13$ times the amount of points total. And since we have to have a whole number of points total, this means that $420 + 28y$ must be divisible by $13$ . Plugging in all the answer choices for $y$ , we find that the only answer that makes $420 + 28y$ divisible by $13$ is $\boxed{11}$",11702214,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_24,1,"In triangle $\triangle ABC$ , point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2$ . Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $\overline{BC}$ and line $\overline{AE}$ . Given that the area of $\triangle ABC$ is $360$ , what is the area of $\triangle EBF$703[asy] unitsize(2cm); pair A,B,C,DD,EE,FF; B = (0,0); C = (3,0);  A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF);  draw(B--DD);dot(A);  label(""$A$"",A,N); dot(B);  label(""$B$"", B,SW);dot(C);  label(""$C$"",C,SE); dot(DD);  label(""$D$"",DD,NE); dot(EE);  label(""$E$"",EE,NW); dot(FF);  label(""$F$"",FF,S); [/asy]704$\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40$","We use the line-segment ratios to infer area ratios and height ratios.705Areas:706$AD:DC = 1:2 \implies AD:AC = 1:3 \implies [ABD] =\frac{[ABC]}{3} = 120$707$BE:BD = 1:2 \text{ (midpoint)}  \implies  [ABE] = \frac{[ABD]}{2} = \frac{120}{2} = 60$708Heights:709Let $h_A$ = height (of altitude) from $\overline{BC}$ to $A$710$AD:DC = 1:2  \implies CD:CA = 2:3  \implies  \text{height } h_D$ from $\overline{BC}$ to $D$ is $\frac{2}{3}h_A$711$BE:BD = 1:2 \text{ (midpoint)}  \implies  \text{height } h_E$ from $\overline{BC}$ to $E$ is $\frac{1}{2} h_D = \frac{1}{2}(\frac{2}{3} h_A) = \frac{1}{3} h_A$712Conclusion:713$\frac{[EBF]} {[ABF]} =  \frac{[EBF]} {[EBF] + [ABE]} =   \frac{[EBF]} {[EBF]+60}$ ,  and also $\frac{[EBF]} {[ABF]} = \frac{h_E}{h_A} = \frac{1}{3}$714So, $\frac{[EBF]} {[EBF] + 60} =  \frac{1}{3}$ , and thus, $[EBF] = \boxed{30}$",30715215,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_24,2,"In triangle $\triangle ABC$ , point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2$ . Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $\overline{BC}$ and line $\overline{AE}$ . Given that the area of $\triangle ABC$ is $360$ , what is the area of $\triangle EBF$716[asy] unitsize(2cm); pair A,B,C,DD,EE,FF; B = (0,0); C = (3,0);  A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF);  draw(B--DD);dot(A);  label(""$A$"",A,N); dot(B);  label(""$B$"", B,SW);dot(C);  label(""$C$"",C,SE); dot(DD);  label(""$D$"",DD,NE); dot(EE);  label(""$E$"",EE,NW); dot(FF);  label(""$F$"",FF,S); [/asy]717$\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40$","Draw $X$ on $\overline{AF}$ such that $\overline{XD}$ is parallel to $\overline{BC}$718Triangles $BEF$ and $EXD$ are similar, and since $BE = ED$ , they are also congruent, and so $XE=EF$ and $XD=BF$719$AC:AD = 3$ implies $\frac{AF}{AX} = 3 = \frac{FC}{XD} = \frac{FC}{BF}$ , so $BC=BF + 3BF = 4BF$ $BF=\frac{BC}{4}$720Since $XE=EF$ $AX = XE = EF$ , and since $AX + XE + EF = AF$ , all of these are equal to $\frac{AF}{3}$ , and so the altitude of triangle $BEF$ is equal to $\frac{1}{3}$ of the altitude of $ABC$721The area of $ABC$ is $360$ , so the area of $\triangle EBF=\frac{1}{3} \cdot \frac{1}{4} \cdot 360=\boxed{30}$",30722216,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_24,3,"In triangle $\triangle ABC$ , point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2$ . Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $\overline{BC}$ and line $\overline{AE}$ . Given that the area of $\triangle ABC$ is $360$ , what is the area of $\triangle EBF$723[asy] unitsize(2cm); pair A,B,C,DD,EE,FF; B = (0,0); C = (3,0);  A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF);  draw(B--DD);dot(A);  label(""$A$"",A,N); dot(B);  label(""$B$"", B,SW);dot(C);  label(""$C$"",C,SE); dot(DD);  label(""$D$"",DD,NE); dot(EE);  label(""$E$"",EE,NW); dot(FF);  label(""$F$"",FF,S); [/asy]724$\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40$","[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(7cm);  real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */  pen dotstyle = black; /* point style */  real xmin = -6.28, xmax = 6.28, ymin = -5.49, ymax = 5.73;  /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451);   /* draw figures */ draw((0.28,2.39)--(-2.8,-1.17), linewidth(2) + wrwrwr);  draw((-2.8,-1.17)--(3.78,-1.05), linewidth(2) + wrwrwr);  draw((3.78,-1.05)--(0.28,2.39), linewidth(2) + wrwrwr);  draw((-2.8,-1.17)--(1.2887445398528459,1.3985482236874887), linewidth(2) + wrwrwr);  draw((0.28,2.39)--(-0.7199623188673492,-1.1320661821070033), linewidth(2) + wrwrwr);  draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr);  draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr);  draw(circle((4.48,-1.28), 0.46818799642878506), linewidth(2) + wrwrwr);  draw(circle((1.98,1.56), 0.46818799642878495), linewidth(2) + wrwrwr);  draw(circle((-3.36,-1.62), 0.46818799642878517), linewidth(2) + wrwrwr);  draw(circle((0.16,0.14), 0.46818799642878495), linewidth(2) + wrwrwr);  draw(circle((-0.74,-1.81), 0.46818799642878495), linewidth(2) + wrwrwr);   /* dots and labels */ dot((0.28,2.39),dotstyle);  label(""$A$"", (0.36,2.59), NE * labelscalefactor);  dot((-2.8,-1.17),dotstyle);  label(""$B$"", (-2.72,-0.97), NE * labelscalefactor);  dot((3.78,-1.05),dotstyle);  label(""$C$"", (3.86,-0.85), NE * labelscalefactor);  dot((1.2887445398528459,1.3985482236874887),dotstyle);  label(""$D$"", (1.36,1.59), NE * labelscalefactor);  dot((-0.7199623188673492,-1.1320661821070033),dotstyle);  label(""$F$"", (-0.64,-0.93), NE * labelscalefactor);  dot((-0.2815567696989588,0.41208536204620183),linewidth(4pt) + dotstyle);  label(""$E$"", (-0.2,0.57), NE * labelscalefactor);  label(""2"", (-0.18,2.81), NE * labelscalefactor,wrwrwr);  label(""1"", (4.4,-1.39), NE * labelscalefactor,wrwrwr);  label(""3"", (1.9,1.45), NE * labelscalefactor,wrwrwr);  label(""3"", (-3.44,-1.73), NE * labelscalefactor,wrwrwr);  label(""6"", (0.08,0.03), NE * labelscalefactor,wrwrwr);  label(""4"", (-0.82,-1.93), NE * labelscalefactor,wrwrwr);  clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);   /* end of picture */ [/asy]725First, when we see the problem, we see ratios, and we see that this triangle basically has no special properties (right, has medians, etc.) and this screams mass points at us.726The triangle we will consider is $\triangle ABC$ (obviously), and we will let $E$ be the center of mass, so that $D$ balances $A$ and $C$ (this is true since $E$ balances $B$ and $D$ , but $E$ also balances $A$ and $B$ and $C$ so $D$ balances $A$ and $C$ ), and $F$ balances $B$ and $C$727We know that $AD:CD=1:2$ and $D$ balances $A$ and $C$ so we assign $2$ to $A$ and $1$ to $C$ . Then, since $D$ balances $A$ and $C$ , we get $D = A + C = 2 + 1 = 3$ (by mass points addition).728Next, since $E$ balances $B$ and $D$ in a ratio of $BE:DE=1:1$ , we know that $B=D=3$ . Similarly, by mass points addition, $E=B+D=3+3=6$729Finally, $F$ balances $B$ and $C$ so $F=B+C=3+1=4$ . We can confirm we have done everything right by noting that $E$ balances $A$ and $F$ , so $E$ should equal $A+F$ , which it does.730Now that our points have weights, we can solve the problem. $BF:FC=1:3$ so $BF:BC=1:4$ so $[ABF]=\frac{1}{4}[ABC]=90$ . Also, $EF:EA=2:4=1:2$ so $EF:AF=1:3$ so $[EBF]=\frac{1}{3}[ABF]=\boxed{30}$",30731217,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_24,4,"In triangle $\triangle ABC$ , point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2$ . Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $\overline{BC}$ and line $\overline{AE}$ . Given that the area of $\triangle ABC$ is $360$ , what is the area of $\triangle EBF$732[asy] unitsize(2cm); pair A,B,C,DD,EE,FF; B = (0,0); C = (3,0);  A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF);  draw(B--DD);dot(A);  label(""$A$"",A,N); dot(B);  label(""$B$"", B,SW);dot(C);  label(""$C$"",C,SE); dot(DD);  label(""$D$"",DD,NE); dot(EE);  label(""$E$"",EE,NW); dot(FF);  label(""$F$"",FF,S); [/asy]733$\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40$","$\frac{BF}{FC}$ is equal to $\frac{\textrm{The area of triangle ABE}}{\textrm{The area of triangle ACE}}$ . The area of triangle $ABE$ is equal to $60$ because it is equal to on half of the area of triangle $ABD$ , which is equal to one-third of the area of triangle $ABC$ , which is $360$ . The area of triangle $ACE$ is the sum of the areas of triangles $AED$ and $CED$ , which is respectively $60$ and $120$ . So, $\frac{BF}{FC}$ is equal to $\frac{60}{180}$ $\frac{1}{3}$ , so the area of triangle $ABF$ is $90$ . That minus the area of triangle $ABE$ is $\boxed{30}$",30734218,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_24,5,"In triangle $\triangle ABC$ , point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2$ . Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $\overline{BC}$ and line $\overline{AE}$ . Given that the area of $\triangle ABC$ is $360$ , what is the area of $\triangle EBF$735[asy] unitsize(2cm); pair A,B,C,DD,EE,FF; B = (0,0); C = (3,0);  A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF);  draw(B--DD);dot(A);  label(""$A$"",A,N); dot(B);  label(""$B$"", B,SW);dot(C);  label(""$C$"",C,SE); dot(DD);  label(""$D$"",DD,NE); dot(EE);  label(""$E$"",EE,NW); dot(FF);  label(""$F$"",FF,S); [/asy]736$\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40$","Extend $\overline{BD}$ to $G$ such that $\overline{AG} \parallel \overline{BC}$ as shown: [asy] size(8cm); pair A, B, C, D, E, F, G; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3);  draw(A--B--C--A--G--B); draw(A--F); label(""$A$"", A, N); label(""$B$"", B, WSW); label(""$C$"", C, ESE); label(""$D$"", D, dir(0)*1.5); label(""$E$"", E, SE); label(""$F$"", F, S); label(""$G$"", G, ENE); [/asy] Then, $\triangle ADG \sim \triangle CDB$ and $\triangle AEG \sim \triangle FEB$ . Since $CD = 2AD$ , triangle $CDB$ has four times the area of triangle $ADG$ . Since $[CDB] = 240$ , we get $[ADG] = 60$737Since $[AED]$ is also $60$ , we have $ED = DG$ because triangles $AED$ and $ADG$ have the same height and same areas and so their bases must be the congruent. Thus, triangle $AEG$ has twice the side lengths and therefore four times the area of triangle $BEF$ , giving $[BEF] = (60+60)/4 = \boxed{30}$",30738219,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_24,7,"In triangle $\triangle ABC$ , point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2$ . Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $\overline{BC}$ and line $\overline{AE}$ . Given that the area of $\triangle ABC$ is $360$ , what is the area of $\triangle EBF$739[asy] unitsize(2cm); pair A,B,C,DD,EE,FF; B = (0,0); C = (3,0);  A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF);  draw(B--DD);dot(A);  label(""$A$"",A,N); dot(B);  label(""$B$"", B,SW);dot(C);  label(""$C$"",C,SE); dot(DD);  label(""$D$"",DD,NE); dot(EE);  label(""$E$"",EE,NW); dot(FF);  label(""$F$"",FF,S); [/asy]740$\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40$","Let $ADB$ be a right triangle, and $BD=CD$741Let $A=(-2\sqrt{30}, 0)$742$B=(0, 4\sqrt{30})$743$C=(4\sqrt{30}, 0)$744$D=(0, 0)$745$E=(0, 2\sqrt{30})$746$F=(\sqrt{30}, 3\sqrt{30})$747The line $\overleftrightarrow{AE}$ can be described with the equation $y=x-2\sqrt{30}$748The line $\overleftrightarrow{BC}$ can be described with $x+y=4\sqrt{30}$749Solving, we get $x=3\sqrt{30}$ and $y=\sqrt{30}$750Now we can find $EF=BF=2\sqrt{15}$751$[\bigtriangleup EBF]=\frac{(2\sqrt{15})^2}{2}=\boxed{30}\blacksquare$",30752220,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_24,8,"In triangle $\triangle ABC$ , point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2$ . Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $\overline{BC}$ and line $\overline{AE}$ . Given that the area of $\triangle ABC$ is $360$ , what is the area of $\triangle EBF$753[asy] unitsize(2cm); pair A,B,C,DD,EE,FF; B = (0,0); C = (3,0);  A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF);  draw(B--DD);dot(A);  label(""$A$"",A,N); dot(B);  label(""$B$"", B,SW);dot(C);  label(""$C$"",C,SE); dot(DD);  label(""$D$"",DD,NE); dot(EE);  label(""$E$"",EE,NW); dot(FF);  label(""$F$"",FF,S); [/asy]754$\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40$","[asy]  /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm);  real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */  pen dotstyle = black; /* point style */  real xmin = -6.61, xmax = 16.13, ymin = -6.4, ymax = 6.42;  /* image dimensions */   /* draw figures */ draw(circle((0,0), 5), linewidth(2));  draw((-4,-3)--(4,3), linewidth(2));  draw((-4,-3)--(0,5), linewidth(2));  draw((0,5)--(4,3), linewidth(2));  draw((12,-1)--(-4,-3), linewidth(2));  draw((0,5)--(0,-5), linewidth(2));  draw((-4,-3)--(0,-5), linewidth(2));  draw((4,3)--(0,2.48), linewidth(2));  draw((4,3)--(12,-1), linewidth(2));  draw((-4,-3)--(4,3), linewidth(2));   /* dots and labels */ dot((0,0),dotstyle);  label(""E"", (0.27,-0.24), NE * labelscalefactor);  dot((-5,0),dotstyle);  dot((-4,-3),dotstyle);  label(""B"", (-4.45,-3.38), NE * labelscalefactor);  dot((4,3),dotstyle);  label(""$D$"", (4.15,3.2), NE * labelscalefactor);  dot((0,5),dotstyle);  label(""A"", (-0.09,5.26), NE * labelscalefactor);  dot((12,-1),dotstyle);  label(""C"", (12.23,-1.24), NE * labelscalefactor);  dot((0,-5),dotstyle);  label(""$G$"", (0.19,-4.82), NE * labelscalefactor);  dot((0,2.48),dotstyle);  label(""I"", (-0.33,2.2), NE * labelscalefactor);  dot((0,0),dotstyle);  label(""E"", (0.27,-0.24), NE * labelscalefactor);  dot((0,-2.5),dotstyle);  label(""F"", (0.23,-2.2), NE * labelscalefactor);  clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);   /* end of picture */ [/asy]755Let $A[\Delta XYZ]$ $\text{Area of Triangle XYZ}$756$A[\Delta ABD]: A[\Delta DBC] :: 1:2 :: 120:240$757$A[\Delta ABE] = A[\Delta AED] = 60$ (the median divides the area of the triangle into two equal parts)758Construction: Draw a circumcircle around $\Delta ABD$ with $BD$ as is diameter. Extend $AF$ to $G$ such that it meets the circle at $G$ . Draw line $BG$759$A[\Delta ABD] = A[\Delta ABG] = 120$ (Since $\square ABGD$ is cyclic)760But $A[\Delta ABE]$ is common in both with an area of 60. So, $A[\Delta AED] = A[\Delta BEG]$761Therefore $A[\Delta AED] \cong A[\Delta BEG]$ (SAS Congruency Theorem).762In $\Delta AED$ , let $DI$ be the median of $\Delta AED$763which means $A[\Delta AID] = 30 = A[\Delta EID]$764Rotate $\Delta DEA$ to meet $D$ at $B$ and $A$ at $G$ $DE$ will fit exactly in $BE$ (both are radii of the circle). From the above solutions, $\frac{AE}{EF} = 2:1$765$AE$ is a radius and $EF$ is half of it implies $EF$ $\frac{radius}{2}$766which means $A[\Delta BEF] \cong A[\Delta DEI]$767Thus, $A[\Delta BEF] = \boxed{30}$",30768221,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_24,9,"In triangle $\triangle ABC$ , point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2$ . Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $\overline{BC}$ and line $\overline{AE}$ . Given that the area of $\triangle ABC$ is $360$ , what is the area of $\triangle EBF$769[asy] unitsize(2cm); pair A,B,C,DD,EE,FF; B = (0,0); C = (3,0);  A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF);  draw(B--DD);dot(A);  label(""$A$"",A,N); dot(B);  label(""$B$"", B,SW);dot(C);  label(""$C$"",C,SE); dot(DD);  label(""$D$"",DD,NE); dot(EE);  label(""$E$"",EE,NW); dot(FF);  label(""$F$"",FF,S); [/asy]770$\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40$","[asy] import geometry; unitsize(2cm); pair A,B,C,DD,EE,FF, M; B = (0,0); C = (3,0); M = (1.45,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF);  draw(B--DD);dot(A);  label(""$A$"",A,N); dot(B);  label(""$B$"", B,SW);dot(C);  label(""$C$"",C,SE); dot(DD);  label(""$D$"",DD,NE); dot(EE);  label(""$E$"",EE,NW); dot(FF);  label(""$F$"",FF,S); draw(EE--M,StickIntervalMarker(1,1)); label(""$M$"",M,S); draw(A--DD,invisible,StickIntervalMarker(1,1)); dot((DD+C)/2); draw(DD--C,invisible,StickIntervalMarker(2,1)); [/asy] Using the ratio of $\overline{AD}$ and $\overline{CD}$ , we find the area of $\triangle ADB$ is $120$ and the area of $\triangle BDC$ is $240$ . Also using the fact that $E$ is the midpoint of $\overline{BD}$ , we know $\triangle ADE = \triangle ABE = 60$ .771Let $M$ be a point such $\overline{EM}$ is parellel to $\overline{CD}$ . We immediatley know that $\triangle BEM \sim BDC$ by $2$ . Using that we can conclude $EM$ has ratio $1$ . Using $\triangle EFM \sim \triangle AFC$ , we get $EF:AE = 1:2$ . Therefore using the fact that $\triangle EBF$ is in $\triangle ABF$ , the area has ratio $\triangle BEF : \triangle ABE=1:2$ and we know $\triangle ABE$ has area $60$ so $\triangle BEF$ is $\boxed{30}$",30772222,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_24,10,"In triangle $\triangle ABC$ , point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2$ . Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $\overline{BC}$ and line $\overline{AE}$ . Given that the area of $\triangle ABC$ is $360$ , what is the area of $\triangle EBF$773[asy] unitsize(2cm); pair A,B,C,DD,EE,FF; B = (0,0); C = (3,0);  A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF);  draw(B--DD);dot(A);  label(""$A$"",A,N); dot(B);  label(""$B$"", B,SW);dot(C);  label(""$C$"",C,SE); dot(DD);  label(""$D$"",DD,NE); dot(EE);  label(""$E$"",EE,NW); dot(FF);  label(""$F$"",FF,S); [/asy]774$\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40$","[asy] unitsize(2cm); pair A,B,C,DD,EE,FF; B = (0,0); C = (3,0);  A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF);  draw(B--DD);dot(A);  label(""$A$"",A,N); dot(B);  label(""$B$"", B,SW);dot(C);  label(""$C$"",C,SE); dot(DD);  label(""$D$"",DD,NE); dot(EE);  label(""$E$"",EE,NW); dot(FF);  label(""$F$"",FF,S); [/asy] By Menelaus's Theorem on triangle $BCD$ , we have \[\dfrac{BF}{FC} \cdot \dfrac{CA}{DA} \cdot \dfrac{DE}{BE} = 3\dfrac{BF}{FC} = 1 \implies \dfrac{BF}{FC} = \dfrac13 \implies \dfrac{BF}{BC} = \dfrac14.\] Therefore, \[[EBF] = \dfrac{BE}{BD}\cdot\dfrac{BF}{BC}\cdot [BCD] = \dfrac12 \cdot \dfrac 14 \cdot \left( \dfrac23 \cdot [ABC]\right) = \boxed{30}.\]",30775223,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_24,11,"In triangle $\triangle ABC$ , point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2$ . Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $\overline{BC}$ and line $\overline{AE}$ . Given that the area of $\triangle ABC$ is $360$ , what is the area of $\triangle EBF$776[asy] unitsize(2cm); pair A,B,C,DD,EE,FF; B = (0,0); C = (3,0);  A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF);  draw(B--DD);dot(A);  label(""$A$"",A,N); dot(B);  label(""$B$"", B,SW);dot(C);  label(""$C$"",C,SE); dot(DD);  label(""$D$"",DD,NE); dot(EE);  label(""$E$"",EE,NW); dot(FF);  label(""$F$"",FF,S); [/asy]777$\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40$","[asy] unitsize(2cm); pair A,B,C,D,E,F,a,b,c,d,e,f; A = (2,3); B = (0,2);  C = (2,0); D = (2/3)*A+(1/3)*C; E = (B+D)/2; F = intersectionpoint(B--C,A--A+2*(E-A)); a = (0,0); b = (1,0); c = (2,1); d = (1,3); e = (0,3); f = (0,1); draw(a--C,dashed); draw(f--c,dashed); draw(e--A,dashed); draw(a--e,dashed); draw(b--d,dashed); draw(A--B--C--cycle); draw(A--F);  draw(B--D); dot(A);  label(""$A$"",A,NE); dot(B);  label(""$B$"",B,dir(180)); dot(C);  label(""$C$"",C,SE); dot(D);  label(""$D$"",D,dir(0)); dot(E);  label(""$E$"",E,SE); dot(F);  label(""$F$"",F,SW); [/asy] Note: If graph paper is unavailable, this solution can still be used by constructing a small grid on a sheet of blank paper. As triangle $ABC$ is loosely defined, we can arrange its points such that the diagram fits nicely on a coordinate plane. By doing so, we can construct it on graph paper and be able to visually determine the relative sizes of the triangles. As point $D$ splits line segment $\overline{AC}$ in a $1:2$ ratio, we draw $\overline{AC}$ as a vertical line segment $3$ units long. Point $D$ is thus $1$ unit below point $A$ and $2$ units above point $C$ . By definition, Point $E$ splits line segment $\overline{BD}$ in a $1:1$ ratio, so we draw $\overline{BD}$ $2$ units long directly left of $D$ and draw $E$ directly between $B$ and $D$ $1$ unit away from both. We then draw line segments $\overline{AB}$ and $\overline{BC}$ . We can easily tell that triangle $ABC$ occupies $3$ square units of space. Constructing line $AE$ and drawing $F$ at the intersection of $AE$ and $BC$ , we can easily see that triangle $EBF$ forms a right triangle occupying $\frac{1}{4}$ of a square unit of space. The ratio of the areas of triangle $EBF$ and triangle $ABC$ is thus $\frac{1}{4}\div3=\frac{1}{12}$ , and since the area of triangle $ABC$ is $360$ , this means that the area of triangle $EBF$ is $\frac{1}{12}\times360=\boxed{30}$",30778224,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_24,12,"In triangle $\triangle ABC$ , point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2$ . Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $\overline{BC}$ and line $\overline{AE}$ . Given that the area of $\triangle ABC$ is $360$ , what is the area of $\triangle EBF$779[asy] unitsize(2cm); pair A,B,C,DD,EE,FF; B = (0,0); C = (3,0);  A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF);  draw(B--DD);dot(A);  label(""$A$"",A,N); dot(B);  label(""$B$"", B,SW);dot(C);  label(""$C$"",C,SE); dot(DD);  label(""$D$"",DD,NE); dot(EE);  label(""$E$"",EE,NW); dot(FF);  label(""$F$"",FF,S); [/asy]780$\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40$","[asy] unitsize(2cm); pair A,B,C,DD,EE,FF,G; B = (0,0); C = (3,0);  A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); G = (1.5,0); draw(A--B--C--cycle); draw(A--FF);  draw(B--DD);  draw(G--DD); label(""$A$"",A,N); label(""$B$"", B,SW);  label(""$C$"",C,SE); label(""$D$"",DD,NE);  label(""$E$"",EE,NW); label(""$F$"",FF,S); label(""$G$"",G,S); [/asy] We know that $AD = \dfrac{1}{3} AC$ , so $[ABD] = \dfrac{1}{3} [ABC] = 120$ . Using the same method, since $BE = \dfrac{1}{2} BD$ $[ABE] = \dfrac{1}{2} [ABD] = 60$ . Next, we draw $G$ on $\overline{BC}$ such that $\overline{DG}$ is parallel to $\overline{AF}$ and create segment $DG$ . We then observe that $\triangle AFC \sim \triangle DGC$ , and since $AD:DC = 1:2$ $FG:GC$ is also equal to $1:2$ . Similarly (no pun intended), $\triangle DBG \sim \triangle EBF$ , and since $BE:ED = 1:1$ $BF:FG$ is also equal to $1:1$ . Combining the information in these two ratios, we find that $BF:FG:GC = 1:1:2$ , or equivalently, $BF = \dfrac{1}{4} BC$ . Thus, $[BFA] = \dfrac{1}{4} [BCA] = 90$ . We already know that $[ABE] = 60$ , so the area of $\triangle EBF$ is $[BFA] - [ABE] = \boxed{30}$",30781225,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_24,13,"In triangle $\triangle ABC$ , point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2$ . Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $\overline{BC}$ and line $\overline{AE}$ . Given that the area of $\triangle ABC$ is $360$ , what is the area of $\triangle EBF$782[asy] unitsize(2cm); pair A,B,C,DD,EE,FF; B = (0,0); C = (3,0);  A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF);  draw(B--DD);dot(A);  label(""$A$"",A,N); dot(B);  label(""$B$"", B,SW);dot(C);  label(""$C$"",C,SE); dot(DD);  label(""$D$"",DD,NE); dot(EE);  label(""$E$"",EE,NW); dot(FF);  label(""$F$"",FF,S); [/asy]783$\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40$","[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0);  draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(""$A$"", A, N); label(""$B$"", B, WSW); label(""$C$"", C, ESE); label(""$D$"", D, dir(0)*1.5); label(""$E$"", E, SSE); label(""$F$"", F, S); label(""$60$"", (A+E+D)/3); label(""$60$"", (A+E+B)/3); label(""$120$"", (D+E+C)/3); label(""$x$"", (B+E+F)/3); label(""$120-x$"", (F+E+C)/3); [/asy] Since $AD:DC=1:2$ thus $\triangle ABD=\frac{1}{3} \cdot 360 = 120.$784Similarly, $\triangle DBC = \frac{2}{3} \cdot 360 = 240.$785Now, since $E$ is a midpoint of $BD$ $\triangle ABE = \triangle AED = 120 \div 2 = 60.$786We can use the fact that $E$ is a midpoint of $BD$ even further. Connect lines $E$ and $C$ so that $\triangle BEC$ and $\triangle DEC$ share 2 sides.787We know that $\triangle BEC=\triangle DEC=240 \div 2 = 120$ since $E$ is a midpoint of $BD.$788Let's label $\triangle BEF$ $x$ . We know that $\triangle EFC$ is $120-x$ since $\triangle BEC = 120.$789Note that with this information now, we can deduct more things that are needed to finish the solution.790Note that $\frac{EF}{AE} = \frac{120-x}{180} = \frac{x}{60}.$ because of triangles $EBF, ABE, AEC,$ and $EFC.$791We want to find $x.$792This is a simple equation, and solving we get $x=\boxed{30}.$",30793226,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_24,14,"In triangle $\triangle ABC$ , point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2$ . Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $\overline{BC}$ and line $\overline{AE}$ . Given that the area of $\triangle ABC$ is $360$ , what is the area of $\triangle EBF$794[asy] unitsize(2cm); pair A,B,C,DD,EE,FF; B = (0,0); C = (3,0);  A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF);  draw(B--DD);dot(A);  label(""$A$"",A,N); dot(B);  label(""$B$"", B,SW);dot(C);  label(""$C$"",C,SE); dot(DD);  label(""$D$"",DD,NE); dot(EE);  label(""$E$"",EE,NW); dot(FF);  label(""$F$"",FF,S); [/asy]795$\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40$","[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0);  draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(""$A$"", A, N); label(""$B$"", B, WSW); label(""$C$"", C, ESE); label(""$D$"", D, dir(0)*1.5); label(""$E$"", E, SSE); label(""$F$"", F, S); label(""$60$"", (A+E+D)/3); label(""$60$"", (A+E+B)/3); label(""$120$"", (D+E+C)/3); [/asy]796Because $AD:DC=1:2$ and $E$ is the midpoint of $BD$ , we know that the areas of $ABE$ and $AED$ are $60$ and the areas of $DEC$ and $EBC$ are $120$ \[\frac{[EBF]}{[EFC]} = \frac{[ABF]}{[AFC]} = \frac{ [ABE]}{[AEC]} = \frac{60}{180}\] $[EBF] = \frac{120}{4} = \boxed{30}$",30797227,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_25,1,Alice has $24$ apples. In how many ways can she share them with Becky and Chris so that each of the three people has at least two apples? $\textbf{(A) }105\qquad\textbf{(B) }114\qquad\textbf{(C) }190\qquad\textbf{(D) }210\qquad\textbf{(E) }380$,"Note: This solution uses the non-negative version for stars and bars. A solution using the positive version of stars is similar (first removing an apple from each person instead of 2).798This method uses the counting method of stars and bars (non-negative version). Since each person must have at least $2$ apples, we can remove $2*3$ apples from the total that need to be sorted. With the remaining $18$ apples, we can use stars and bars to determine the number of possibilities. Assume there are $18$ stars in a row, and $2$ bars, which will be placed to separate the stars into groups of $3$ . In total, there are $18$ spaces for stars $+ 2$ spaces for bars, for a total of $20$ spaces. We can now do $20 \choose 2$ . This is because if we choose distinct $2$ spots for the bars to be placed, each combo of $3$ groups will be different, and all apples will add up to $18$ . We can also do this because the apples are indistinguishable. $20 \choose 2$ is $190$ , therefore the answer is $\boxed{190}$",190799228,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_25,2,Alice has $24$ apples. In how many ways can she share them with Becky and Chris so that each of the three people has at least two apples? $\textbf{(A) }105\qquad\textbf{(B) }114\qquad\textbf{(C) }190\qquad\textbf{(D) }210\qquad\textbf{(E) }380$,"Consider an unordered triple $(a,b,c)$ where $a+b+c=24$ and $a,b,c$ are not necessarily distinct. Then, we will either have $1$ $3$ , or $6$ distinguishable ways to assign $a$ $b$ , and $c$ to Alice, Becky, and Chris. Thus, our answer will be $x+3y+6z$ for some nonnegative integers $x,y,z$ . Notice that we only have $1$ way to assign the numbers $a,b,c$ to Alice, Becky, and Chris when $a=b=c$ . As this only happens $1$ way ( $a=b=c=8$ ), our answer is $1+3y+6z$ for some $y,z$ . Finally, notice that this implies the answer is $1$ mod $3$ . The only answer choice that satisfies this is $\boxed{190}$",190800229,https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_25,3,Alice has $24$ apples. In how many ways can she share them with Becky and Chris so that each of the three people has at least two apples? $\textbf{(A) }105\qquad\textbf{(B) }114\qquad\textbf{(C) }190\qquad\textbf{(D) }210\qquad\textbf{(E) }380$,"Since each person needs to have at least two apples, we can simply give each person two, leaving $24 - 2\times3=18$ apples. For the remaining apples, if Alice is going to have $a$ apples, Becky is going to have $b$ apples, and Chris is going to have $c$ apples, we have indeterminate equation $a+b+c=18$ . Currently, we can see that $0 \leq a\leq 18$ where $a$ is an integer, and when $a$ equals any number in the range, there will be $18-a+1=19-a$ sets of values for $b$ and $c$ . Thus, there are $19 + 18 + 17 + \cdots + 1 = \boxed{190}$ possible sets of values in total.",190801230,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_1,1,"An amusement park has a collection of scale models, with a ratio of $1: 20$ , of buildings and other sights from around the country. The height of the United States Capitol is $289$ feet. What is the height in feet of its duplicate to the nearest whole number?802$\textbf{(A) }14\qquad\textbf{(B) }15\qquad\textbf{(C) }16\qquad\textbf{(D) }18\qquad\textbf{(E) }20$","You can see that since the ratio of real building's heights to the model building's height is $1:20$ . We also know that the U.S Capitol is $289$ feet in real life, so to find the height of the model, we divide by 20. That gives us $14.45$ which rounds to 14. Therefore, to the nearest whole number, the duplicate is $\boxed{14}$",14803231,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_1,3,"An amusement park has a collection of scale models, with a ratio of $1: 20$ , of buildings and other sights from around the country. The height of the United States Capitol is $289$ feet. What is the height in feet of its duplicate to the nearest whole number?804$\textbf{(A) }14\qquad\textbf{(B) }15\qquad\textbf{(C) }16\qquad\textbf{(D) }18\qquad\textbf{(E) }20$","We know that $20 \cdot 14 = 280 ,$ and that $20 \cdot 15 = 300 .$ These are the multiples of $20$ around $289 ,$ and the closest one of those is $280.$ Therefore, the answer is $\dfrac {280} {20} = \boxed{14} .$",14805232,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_2,1,"What is the value of the product806\[\left(1+\frac{1}{1}\right)\cdot\left(1+\frac{1}{2}\right)\cdot\left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{4}\right)\cdot\left(1+\frac{1}{5}\right)\cdot\left(1+\frac{1}{6}\right)?\]807$\textbf{(A) }\frac{7}{6}\qquad\textbf{(B) }\frac{4}{3}\qquad\textbf{(C) }\frac{7}{2}\qquad\textbf{(D) }7\qquad\textbf{(E) }8$","By adding up the numbers in each of the $6$ parentheses, we get:808$\frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5} \cdot \frac{7}{6}$809Using telescoping, most of the terms cancel out diagonally. We are left with $\frac{7}{1}$ which is equivalent to $7$ . Thus, the answer would be $\boxed{7}$",7810233,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_4,1,"The twelve-sided figure shown has been drawn on $1 \text{ cm}\times 1 \text{ cm}$ graph paper. What is the area of the figure in $\text{cm}^2$811[asy] unitsize(8mm); for (int i=0; i<7; ++i) {   draw((i,0)--(i,7),gray);   draw((0,i+1)--(7,i+1),gray); } draw((1,3)--(2,4)--(2,5)--(3,6)--(4,5)--(5,5)--(6,4)--(5,3)--(5,2)--(4,1)--(3,2)--(2,2)--cycle,black+2bp); [/asy]812$\textbf{(A) } 12 \qquad \textbf{(B) } 12.5 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 13.5 \qquad \textbf{(E) } 14$","We count $3 \cdot 3=9$ unit squares in the middle, and $8$ small triangles, which gives 4 rectangles each with an area of $1$ . Thus, the answer is $9+4=\boxed{13}$",13813234,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_4,2,"The twelve-sided figure shown has been drawn on $1 \text{ cm}\times 1 \text{ cm}$ graph paper. What is the area of the figure in $\text{cm}^2$814[asy] unitsize(8mm); for (int i=0; i<7; ++i) {   draw((i,0)--(i,7),gray);   draw((0,i+1)--(7,i+1),gray); } draw((1,3)--(2,4)--(2,5)--(3,6)--(4,5)--(5,5)--(6,4)--(5,3)--(5,2)--(4,1)--(3,2)--(2,2)--cycle,black+2bp); [/asy]815$\textbf{(A) } 12 \qquad \textbf{(B) } 12.5 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 13.5 \qquad \textbf{(E) } 14$","We can see here that there are $9$ total squares in the middle. We also see that the triangles that make the corners of the shape have an area half the squares' area. Then, we can easily find that each corner has an area of one square and there are $4$ corners so we add that to the original 9 squares to get $9+4=\boxed{13}$ . That is how I did it.",13816235,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_4,3,"The twelve-sided figure shown has been drawn on $1 \text{ cm}\times 1 \text{ cm}$ graph paper. What is the area of the figure in $\text{cm}^2$817[asy] unitsize(8mm); for (int i=0; i<7; ++i) {   draw((i,0)--(i,7),gray);   draw((0,i+1)--(7,i+1),gray); } draw((1,3)--(2,4)--(2,5)--(3,6)--(4,5)--(5,5)--(6,4)--(5,3)--(5,2)--(4,1)--(3,2)--(2,2)--cycle,black+2bp); [/asy]818$\textbf{(A) } 12 \qquad \textbf{(B) } 12.5 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 13.5 \qquad \textbf{(E) } 14$","We can apply Pick's Theorem here. There are $8$ lattice points, and $12$ lattice points on the boundary. Then,819\[8 + 12 \div 2 - 1 = \boxed{13}.\]",13820236,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_5,1,"What is the value of $1+3+5+\cdots+2017+2019-2-4-6-\cdots-2016-2018$821$\textbf{(A) }-1010\qquad\textbf{(B) }-1009\qquad\textbf{(C) }1008\qquad\textbf{(D) }1009\qquad \textbf{(E) }1010$","Rearranging the terms, we get $(1-2)+(3-4)+(5-6)+...(2017-2018)+2019$ , and our answer is $-1009+2019=\boxed{1010}$",10822237,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_5,2,"What is the value of $1+3+5+\cdots+2017+2019-2-4-6-\cdots-2016-2018$823$\textbf{(A) }-1010\qquad\textbf{(B) }-1009\qquad\textbf{(C) }1008\qquad\textbf{(D) }1009\qquad \textbf{(E) }1010$","We can see that the last numbers of each of the sets (even numbers and odd numbers) have a difference of two. So, do the second last ones and so on. Now, all we need to find is the number of integers in any of the sets (I chose even) to get $\boxed{1010}$",10824238,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_5,3,"What is the value of $1+3+5+\cdots+2017+2019-2-4-6-\cdots-2016-2018$825$\textbf{(A) }-1010\qquad\textbf{(B) }-1009\qquad\textbf{(C) }1008\qquad\textbf{(D) }1009\qquad \textbf{(E) }1010$","It is similar to the Solution 1: 826Rearranging the terms, we get $1+(3-2)+(5-4)+(6-5)...(2017-2016)+(2019-2018)$ , and our answer is $1+1009=\boxed{1010}$",10827239,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_5,4,"What is the value of $1+3+5+\cdots+2017+2019-2-4-6-\cdots-2016-2018$828$\textbf{(A) }-1010\qquad\textbf{(B) }-1009\qquad\textbf{(C) }1008\qquad\textbf{(D) }1009\qquad \textbf{(E) }1010$","Note that the sum of consecutive odd numbers can be expressed as a square, namely $1+3+5+7+...+2017+2019 = 1010^2$ . We can modify the negative numbers in the same way by adding 1 to each negative term, factoring a negative sign, and accounting for the extra 1's by subtracting 1009. We then have $1010^2-1009^2-1009$ . Using difference of squares, we obtain $(1010+1009)(1010-1009)-1009 = 2019-1009 = \boxed{1010}$",10829240,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_6,1,"On a trip to the beach, Anh traveled 50 miles on the highway and 10 miles on a coastal access road. He drove three times as fast on the highway as on the coastal road. If Anh spent 30 minutes driving on the coastal road, how many minutes did his entire trip take?830[mathjax]\textbf{(A) }50\qquad\textbf{(B) }70\qquad\textbf{(C) }80\qquad\textbf{(D) }90\qquad \textbf{(E) }100[/mathjax]","Since Anh spends half an hour to drive 10 miles on the coastal road, his speed is [mathjax]r=\dfrac dt=\dfrac{10}{0.5}=20[/mathjax] mph. His speed on the highway then is [mathjax]60[/mathjax] mph. He drives [mathjax]50[/mathjax] miles, so he drives for [mathjax]\dfrac{5}{6}[/mathjax] hours, which is equal to [mathjax]50[/mathjax] minutes (Note that [mathjax]60[/mathjax] miles per hour is the same as [mathjax]1[/mathjax] mile per minute). The total amount of minutes spent on his trip is [mathjax]30+50\implies \boxed{80}[/mathjax].",80831241,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_6,2,"On a trip to the beach, Anh traveled 50 miles on the highway and 10 miles on a coastal access road. He drove three times as fast on the highway as on the coastal road. If Anh spent 30 minutes driving on the coastal road, how many minutes did his entire trip take?832[mathjax]\textbf{(A) }50\qquad\textbf{(B) }70\qquad\textbf{(C) }80\qquad\textbf{(D) }90\qquad \textbf{(E) }100[/mathjax]","Since Anh drives [mathjax]3[/mathjax] times as fast on the highway, it takes him [mathjax]\dfrac{1}{3}[/mathjax] of the time to drive [mathjax]10[/mathjax] miles on the highway than on the coastal road. [mathjax]\dfrac{1}{3}[/mathjax] of [mathjax]30[/mathjax] is [mathjax]10[/mathjax], and since he drives [mathjax]50[/mathjax] miles on the highway, we multiply [mathjax]10[/mathjax] by [mathjax]5[/mathjax] to get [mathjax]50[/mathjax]. This means it took him [mathjax]50[/mathjax] minutes to drive on the highway, and if we add the [mathjax]30[/mathjax] minutes it took for him to drive on the coastal road, we would get [mathjax]\boxed{80}[/mathjax].",80833242,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_7,1,"The $5$ -digit number $\underline{2}$ $\underline{0}$ $\underline{1}$ $\underline{8}$ $\underline{U}$ is divisible by $9$ . What is the remainder when this number is divided by $8$834$\textbf{(A) }1\qquad\textbf{(B) }3\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7$","We use the property that the digits of a number must sum to a multiple of $9$ if it are divisible by $9$ . This means $2+0+1+8+U$ must be divisible by $9$ . The only possible value for $U$ then must be $7$ . Since we are looking for the remainder when divided by $8$ , we can ignore the thousands. The remainder when $187$ is divided by $8$ is $\boxed{3}$",3835243,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_9,1,"Monica is tiling the floor of her 12-foot by 16-foot living room. She plans to place one-foot by one-foot square tiles to form a border along the edges of the room and to fill in the rest of the floor with two-foot by two-foot square tiles. How many tiles will she use?836$\textbf{(A) }48\qquad\textbf{(B) }87\qquad\textbf{(C) }89\qquad\textbf{(D) }96\qquad \textbf{(E) }120$","She will place $(12\cdot2)+(14\cdot2)=52$ tiles around the border. For the inner part of the room, we have $10\cdot14=140$ square feet. Each tile takes up $4$ square feet, so he will use $\frac{140}{4}=35$ tiles for the inner part of the room. Thus, the answer is $52+35= \boxed{87}$",87837244,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_9,2,"Monica is tiling the floor of her 12-foot by 16-foot living room. She plans to place one-foot by one-foot square tiles to form a border along the edges of the room and to fill in the rest of the floor with two-foot by two-foot square tiles. How many tiles will she use?838$\textbf{(A) }48\qquad\textbf{(B) }87\qquad\textbf{(C) }89\qquad\textbf{(D) }96\qquad \textbf{(E) }120$","The area around the border: $(12 \cdot 2) + (14 \cdot 2) = 52$ . The area of tiles around the border: $1 \cdot 1 = 1$ . Therefore, $\frac{52}{1} = 52$ is the number of tiles around the border.839The inner part will have $(12 - 2)(16 - 2) = 140$ . The area of those tiles are $2 \cdot 2 = 4$ $\frac{140}{4} = 35$ is the amount of tiles for the inner part. So, $52 + 35 = \boxed{87}$",87840245,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_13,1,"Laila took five math tests, each worth a maximum of 100 points. Laila's score on each test was an integer between 0 and 100, inclusive. Laila received the same score on the first four tests, and she received a higher score on the last test. Her average score on the five tests was 82. How many values are possible for Laila's score on the last test?841$\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad \textbf{(E) }18$","Say Laila gets a value of $x$ on her first 4 tests, and a value of $y$ on her last test. Thus, $4x+y=410.$842The value $y$ has to be greater than $82$ , because otherwise she would receive the same score on her last test. Additionally, the greatest value for $y$ is $98$ (as $y=100$ would make $x$ as a decimal), so therefore, the greatest value $x$ can be is $98$ . As a result, only $4$ numbers work, $86, 90, 94$ and $98$ . Thus, the answer is $\boxed{4}$",4843246,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_13,2,"Laila took five math tests, each worth a maximum of 100 points. Laila's score on each test was an integer between 0 and 100, inclusive. Laila received the same score on the first four tests, and she received a higher score on the last test. Her average score on the five tests was 82. How many values are possible for Laila's score on the last test?844$\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad \textbf{(E) }18$","The average point is $82$ leads us to suppose that Laila got all $82$ points for the tests. We know that Laila got the same points in the first four tests and they are all lower than the last test. Let the first four tests is $81$ points, then the last tests should be $86$ points to keep the average point is fixed. The possible points are $86$ $90$ $94$ $98$ . The answer is $\boxed{4}$",4845247,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_14,1,"Let $N$ be the greatest five-digit number whose digits have a product of $120$ . What is the sum of the digits of $N$846$\textbf{(A) }15\qquad\textbf{(B) }16\qquad\textbf{(C) }17\qquad\textbf{(D) }18\qquad\textbf{(E) }20$","If we start off with the first digit, we know that it can't be $9$ since $9$ is not a factor of $120$ . We go down to the digit $8$ , which does work since it is a factor of $120$ . Now, we have to know what digits will take up the remaining four spots. To find this result, just divide $\frac{120}{8}=15$ . The next place can be $5$ , as it is the largest factor, aside from $15$ . Consequently, our next three values will be $3,1$ and $1$ if we use the same logic. Therefore, our five-digit number is $85311$ , so the sum is $8+5+3+1+1=18\implies \boxed{18}$",18847248,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_14,2,"Let $N$ be the greatest five-digit number whose digits have a product of $120$ . What is the sum of the digits of $N$848$\textbf{(A) }15\qquad\textbf{(B) }16\qquad\textbf{(C) }17\qquad\textbf{(D) }18\qquad\textbf{(E) }20$","120 is 5!, so we have 5,4,3,2,1. Now look for the largest digit which you multiple numbers.849$(5)(4)(3)(2)(1) = 120$ Making the greatest integer,850$(5)(4 \cdot 2)(3)\left(\frac{2}{2}\right)(1)$ $= (5)(8)(3)(1)(1) =120$8518 is the largest value and will go in the front.852We can express the number as $85311$853$8+5+3+1+1=\boxed{18}$",18854249,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_15,1,"In the diagram below, a diameter of each of the two smaller circles is a radius of the larger circle. If the two smaller circles have a combined area of $1$ square unit, then what is the area of the shaded region, in square units?855[asy] size(4cm); filldraw(scale(2)*unitcircle,gray,black); filldraw(shift(-1,0)*unitcircle,white,black); filldraw(shift(1,0)*unitcircle,white,black); [/asy]856$\textbf{(A) } \frac{1}{4} \qquad \textbf{(B) } \frac{1}{3} \qquad \textbf{(C) } \frac{1}{2} \qquad \textbf{(D) } 1 \qquad \textbf{(E) } \frac{\pi}{2}$","Let the radius of the large circle be $R$ . Then, the radius of the smaller circles are $\frac R2$ . The areas of the circles are directly proportional to the square of the radii, so the ratio of the area of the small circle to the large one is $\frac 14$ . This means the combined area of the 2 smaller circles is half of the larger circle, and therefore the shaded region is equal to the combined area of the 2 smaller circles, which is $\boxed{1}$",1857250,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_15,2,"In the diagram below, a diameter of each of the two smaller circles is a radius of the larger circle. If the two smaller circles have a combined area of $1$ square unit, then what is the area of the shaded region, in square units?858[asy] size(4cm); filldraw(scale(2)*unitcircle,gray,black); filldraw(shift(-1,0)*unitcircle,white,black); filldraw(shift(1,0)*unitcircle,white,black); [/asy]859$\textbf{(A) } \frac{1}{4} \qquad \textbf{(B) } \frac{1}{3} \qquad \textbf{(C) } \frac{1}{2} \qquad \textbf{(D) } 1 \qquad \textbf{(E) } \frac{\pi}{2}$","Let the radius of the two smaller circles be $r$ . It follows that the area of one of the smaller circles is ${\pi}r^2$ . Thus, the area of the two inner circles combined would evaluate to $2{\pi}r^2$ which is $1$ . Since the radius of the bigger circle is two times that of the smaller circles (the diameter), the radius of the larger circle in terms of $r$ would be $2r$ . The area of the larger circle would come to $(2r)^2{\pi} = 4{\pi}r^2$860Subtracting the area of the smaller circles from that of the larger circle (since that would be the shaded region), we have \[4{\pi}r^2 - 2{\pi}r^2 = 2{\pi}r^2 = 1.\]861Therefore, the area of the shaded region is $\boxed{1}$",1862251,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_16,1,"Professor Chang has nine different language books lined up on a bookshelf: two Arabic, three German, and four Spanish. How many ways are there to arrange the nine books on the shelf keeping the Arabic books together and keeping the Spanish books together?863$\textbf{(A) }1440\qquad\textbf{(B) }2880\qquad\textbf{(C) }5760\qquad\textbf{(D) }182,440\qquad \textbf{(E) }362,880$","Since the Arabic books and Spanish books have to be kept together, we can treat them both as just one book. That means we're trying to find the number of ways you can arrange one Arabic book, one Spanish book, and three German books, which is just $5$ factorial. Now, we multiply this product by $2!$ because there are $2!$ ways to arrange the Arabic books within themselves, and $4!$ ways to arrange the Spanish books within themselves. Multiplying all these together, we have $2! \cdot 4! \cdot 5!=\boxed{5760}$",760864252,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_17,1,"Bella begins to walk from her house toward her friend Ella's house. At the same time, Ella begins to ride her bicycle toward Bella's house. They each maintain a constant speed, and Ella rides $5$ times as fast as Bella walks. The distance between their houses is $2$ miles, which is $10,560$ feet, and Bella covers $2 \tfrac{1}{2}$ feet with each step. How many steps will Bella take by the time she meets Ella?865$\textbf{(A) }704\qquad\textbf{(B) }845\qquad\textbf{(C) }1056\qquad\textbf{(D) }1760\qquad \textbf{(E) }3520$","Every 10 feet Bella goes, Ella goes 50 feet, which means a total of 60 feet. They need to travel that 60 feet $10560\div60=176$ times to travel the entire 2 miles. Since Bella goes 10 feet 176 times, this means that she travels a total of 1760 feet. And since she walks 2.5 feet each step, $1760\div2.5=\boxed{704}$",704866253,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_17,2,"Bella begins to walk from her house toward her friend Ella's house. At the same time, Ella begins to ride her bicycle toward Bella's house. They each maintain a constant speed, and Ella rides $5$ times as fast as Bella walks. The distance between their houses is $2$ miles, which is $10,560$ feet, and Bella covers $2 \tfrac{1}{2}$ feet with each step. How many steps will Bella take by the time she meets Ella?867$\textbf{(A) }704\qquad\textbf{(B) }845\qquad\textbf{(C) }1056\qquad\textbf{(D) }1760\qquad \textbf{(E) }3520$","We know that Bella goes 2.5 feet per step and since Ella rides 5 times faster than Bella she must go 12.5 feet on her bike for every step of Bella's. For Bella, it takes 4,224 steps, and for Ella, it takes 1/5th those steps since Ella goes 5 times faster than Bella, taking her 844.8 steps. The number of steps where they meet therefore must be less than 844.8. The only answer choice less than it is $\boxed{704}$",704868254,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_17,3,"Bella begins to walk from her house toward her friend Ella's house. At the same time, Ella begins to ride her bicycle toward Bella's house. They each maintain a constant speed, and Ella rides $5$ times as fast as Bella walks. The distance between their houses is $2$ miles, which is $10,560$ feet, and Bella covers $2 \tfrac{1}{2}$ feet with each step. How many steps will Bella take by the time she meets Ella?869$\textbf{(A) }704\qquad\textbf{(B) }845\qquad\textbf{(C) }1056\qquad\textbf{(D) }1760\qquad \textbf{(E) }3520$","We can turn $2 \tfrac{1}{2}$ into an improper fraction. It will then become 5/2. Since Ella bikes 5 times faster, we multiply 5/2 by 5 to get 25/2. Then we add 5/2 to it in order to find the distance they walk and bike together in total. After adding, you should get 30/2 which is equal to 15. This means that after 15 times, they will meet. So you have to divide 10,560 by 15. The answer should be $\boxed{704}$",704870255,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_18,1,"How many positive factors does $23,232$ have?871$\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }42$","We can first find the prime factorization of $23,232$ , which is $2^6\cdot3^1\cdot11^2$ . Now, we add one to our powers and multiply. Therefore, the answer is $(6+1)\cdot(1+1)\cdot(2+1)=7\cdot2\cdot3=\boxed{42}$",42872256,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_18,2,"How many positive factors does $23,232$ have?873$\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }42$","Observe that $69696$ $264^2$ , so this is $\frac{1}{3}$ of $264^2$ which is $88 \cdot 264 = 11^2 \cdot 8^2 \cdot 3 = 11^2 \cdot 2^6 \cdot 3$ , which has $3 \cdot 7 \cdot 2 = 42$ factors. The answer is $\boxed{42}$",42874257,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_19,1,"In a sign pyramid a cell gets a ""+"" if the two cells below it have the same sign, and it gets a ""-"" if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a ""+"" at the top of the pyramid?875[asy] unitsize(2cm); path box = (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--cycle; draw(box); label(""$+$"",(0,0)); draw(shift(1,0)*box); label(""$-$"",(1,0)); draw(shift(2,0)*box); label(""$+$"",(2,0)); draw(shift(3,0)*box); label(""$-$"",(3,0)); draw(shift(0.5,0.4)*box); label(""$-$"",(0.5,0.4)); draw(shift(1.5,0.4)*box); label(""$-$"",(1.5,0.4)); draw(shift(2.5,0.4)*box); label(""$-$"",(2.5,0.4)); draw(shift(1,0.8)*box); label(""$+$"",(1,0.8)); draw(shift(2,0.8)*box); label(""$+$"",(2,0.8)); draw(shift(1.5,1.2)*box); label(""$+$"",(1.5,1.2)); [/asy]876$\textbf{(A) } 2 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16$","You could just make out all of the patterns that make the top positive. In this case, you would have the following patterns:877+−−+, −++−, −−−−, ++++, −+−+, +−+−, ++−−, −−++. There are 8 patterns and so the answer is $\boxed{8}$",8878258,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_19,2,"In a sign pyramid a cell gets a ""+"" if the two cells below it have the same sign, and it gets a ""-"" if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a ""+"" at the top of the pyramid?879[asy] unitsize(2cm); path box = (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--cycle; draw(box); label(""$+$"",(0,0)); draw(shift(1,0)*box); label(""$-$"",(1,0)); draw(shift(2,0)*box); label(""$+$"",(2,0)); draw(shift(3,0)*box); label(""$-$"",(3,0)); draw(shift(0.5,0.4)*box); label(""$-$"",(0.5,0.4)); draw(shift(1.5,0.4)*box); label(""$-$"",(1.5,0.4)); draw(shift(2.5,0.4)*box); label(""$-$"",(2.5,0.4)); draw(shift(1,0.8)*box); label(""$+$"",(1,0.8)); draw(shift(2,0.8)*box); label(""$+$"",(2,0.8)); draw(shift(1.5,1.2)*box); label(""$+$"",(1.5,1.2)); [/asy]880$\textbf{(A) } 2 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16$","The top box is fixed by the problem.881Choose the left 3 bottom-row boxes freely. There are $2^3=8$ ways.882Then the left 2 boxes on the row above are determined.883Then the left 1 box on the row above that is determined884Then the right 1 box on that row is determined.885Then the right 1 box on the row below is determined.886Then the right 1 box on the bottom row is determined, completing the diagram.887So the answer is $\boxed{8}$",8888259,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_19,3,"In a sign pyramid a cell gets a ""+"" if the two cells below it have the same sign, and it gets a ""-"" if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a ""+"" at the top of the pyramid?889[asy] unitsize(2cm); path box = (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--cycle; draw(box); label(""$+$"",(0,0)); draw(shift(1,0)*box); label(""$-$"",(1,0)); draw(shift(2,0)*box); label(""$+$"",(2,0)); draw(shift(3,0)*box); label(""$-$"",(3,0)); draw(shift(0.5,0.4)*box); label(""$-$"",(0.5,0.4)); draw(shift(1.5,0.4)*box); label(""$-$"",(1.5,0.4)); draw(shift(2.5,0.4)*box); label(""$-$"",(2.5,0.4)); draw(shift(1,0.8)*box); label(""$+$"",(1,0.8)); draw(shift(2,0.8)*box); label(""$+$"",(2,0.8)); draw(shift(1.5,1.2)*box); label(""$+$"",(1.5,1.2)); [/asy]890$\textbf{(A) } 2 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16$","Let the plus sign represent 1 and the negative sign represent -1.891The four numbers on the bottom are $a$ $b$ $c$ , and $d$ , which are either 1 or -1.892[asy] unitsize(2cm); path box = (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--cycle; draw(box); label(""$a$"",(0,0)); draw(shift(1,0)*box); label(""$b$"",(1,0)); draw(shift(2,0)*box); label(""$c$"",(2,0)); draw(shift(3,0)*box); label(""$d$"",(3,0)); draw(shift(0.5,0.4)*box); label(""$ab$"",(0.5,0.4)); draw(shift(1.5,0.4)*box); label(""$bc$"",(1.5,0.4)); draw(shift(2.5,0.4)*box); label(""$cd$"",(2.5,0.4)); draw(shift(1,0.8)*box); label(""$ab^2c$"",(1,0.8)); draw(shift(2,0.8)*box); label(""$bc^2d$"",(2,0.8)); draw(shift(1.5,1.2)*box); label(""$ab^3c^3d$"",(1.5,1.2)); [/asy]893Which means $ab^3c^3d$ = 1. Since $b$ and $c$ are either 1 or -1, $b^3 = b$ and $c^3 = c$ . This shows that $abcd$ = 1.894Therefore either $a$ $b$ $c$ , and $d$ are all positive or negative, or 2 are positive and 2 are negative.895There are 2 ways where $a$ $b$ $c$ , and $d$ are 1 (1, 1, 1, 1) and (-1, -1, -1, -1)896There are 6 ways where 2 variables are positive and 2 are negative: (1, 1, -1, -1), (1, -1, 1, -1), (-1, 1, 1, -1), (-1, -1, 1, 1), (-1, 1, -1, 1), and (-1, -1, 1, 1).897So the answer is $\boxed{8}$",8898260,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_19,4,"In a sign pyramid a cell gets a ""+"" if the two cells below it have the same sign, and it gets a ""-"" if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a ""+"" at the top of the pyramid?899[asy] unitsize(2cm); path box = (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--cycle; draw(box); label(""$+$"",(0,0)); draw(shift(1,0)*box); label(""$-$"",(1,0)); draw(shift(2,0)*box); label(""$+$"",(2,0)); draw(shift(3,0)*box); label(""$-$"",(3,0)); draw(shift(0.5,0.4)*box); label(""$-$"",(0.5,0.4)); draw(shift(1.5,0.4)*box); label(""$-$"",(1.5,0.4)); draw(shift(2.5,0.4)*box); label(""$-$"",(2.5,0.4)); draw(shift(1,0.8)*box); label(""$+$"",(1,0.8)); draw(shift(2,0.8)*box); label(""$+$"",(2,0.8)); draw(shift(1.5,1.2)*box); label(""$+$"",(1.5,1.2)); [/asy]900$\textbf{(A) } 2 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16$","The pyramid is built on the basic 3 blocks pattern: one above and two below. The basic pattern have four possible symbols and half of them have a $+$ on the above, half of them have a $-$ above. So, For the lowest layer with $4$ blocks, there are $2^4=16$ possible combination and half of them will lead a $+$ (or $-$ ) on the top. The answer is $16/2=\boxed{8}$",8901261,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_21,1,"How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?902$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$","Looking at the values, we notice that $11-7=4$ $9-5=4$ and $6-2=4$ . This means we are looking for a value that is four less than a multiple of $11$ $9$ , and $6$ . The least common multiple of these numbers is $11\cdot3^{2}\cdot2=198$ , so the numbers that fulfill this can be written as $198k-4$ , where $k$ is a positive integer.  This value is only a three-digit integer when $k$ is $1, 2, 3, 4$ or $5$ , which gives $194, 392, 590, 788,$ and $986$ respectively. Thus, we have $5$ values, so our answer is $\boxed{5}$",5903262,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_21,2,"How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?904$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$","Let us create the equations: $6x+2 = 9y+5 = 11z+7$ , and we know $100 \leq 11z+7 <1000$ , it gives us $9 \leq z \leq 90$ , which is the range of the value of z. Because of $6x+2=11z+7$ , then $6x=11z+5=6z+5(z+1)$ , so $(z+1)$ must be a mutiple of 6. Because of $9y+5=11z+7$ , then $9y=11z+2=9z+2(z+1)$ , so $(z+1)$ must also be a mutiple of $9$ . Hence, the value of $(z+1)$ must be a common multiple of $6$ and $9$ , which means multiples of $18 (LCM \text{ of }\ 6, 9)$ . So, let's say $z+1 = 18p$ ; then, $9 \leq z = 18p-1 \leq 90$ ,  so $1 \leq p \leq 91/18\ or \ 1 \leq p \leq 5$ . Thus, the answer is $\boxed{5}$",5905263,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_21,3,"How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?906$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$","By the Chinese Remainder Theorem , we have that all solutions are in the form $x=198k+194$ where $k\in \mathbb{Z}.$ Counting the number of values, we get $\boxed{5}.$",5907264,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_21,4,"How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?908$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$","We can use modular arithmetic. Set up the equations: $x \equiv 2 \mod 6,$ $x \equiv 5 \mod 9,$ and $x \equiv 7 \mod 11.$ These equations can also be written as $x+4 \equiv 0 \mod 6,$ $x+4 \equiv 0 \mod 9,$ and $x+4 \equiv 0 \mod 11.$ Since $x+4$ is congruent to numbers $6, 9,$ and $11,$ then it must also be congruent to their LCM. Thus, $x+4 \equiv 0 \mod 198,$ since 198 is the LCM of $6, 9,$ and $11.$ Since these numbers have to be three digits, they can only be $194, 392, 590, 788,$ and $986.$ This gives us the answer of $\boxed{5}.$",5909265,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_21,5,"How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?910$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$","Let $N$ be the three digit positive integer. $N = 6a + 2 = 9b + 5 = 11c + 7$ . Then, we add four to all sides and write $N + 4 = 6(a+1) = 9(b+1) = 11(c+1)$ . Now, we know that $N + 4$ is divisible by 6, 9, and 11. The LCM of 6, 9, and 11 is equal to 198, so $N = 198k - 4$ . From this, we can figure out that $N$ can be 5 different three digit numbers -- $194, 392, 590, 788,$ and $986$ $\therefore$ , the answer is $\boxed{5}.$",5911266,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_22,1,"Point $E$ is the midpoint of side $\overline{CD}$ in square $ABCD,$ and $\overline{BE}$ meets diagonal $\overline{AC}$ at $F.$ The area of quadrilateral $AFED$ is $45.$ What is the area of $ABCD?$912[asy] size(5cm); draw((0,0)--(6,0)--(6,6)--(0,6)--cycle); draw((0,6)--(6,0)); draw((3,0)--(6,6)); label(""$A$"",(0,6),NW); label(""$B$"",(6,6),NE); label(""$C$"",(6,0),SE); label(""$D$"",(0,0),SW); label(""$E$"",(3,0),S); label(""$F$"",(4,2),E); [/asy]913$\textbf{(A) } 100 \qquad \textbf{(B) } 108 \qquad \textbf{(C) } 120 \qquad \textbf{(D) } 135 \qquad \textbf{(E) } 144$","Let the area of $\triangle CEF$ be $x$ . Thus, the area of triangle $\triangle ACD$ is $45+x$ and the area of the square is $2(45+x) = 90+2x$914By AA similarity, $\triangle CEF \sim \triangle ABF$ with a 1:2 ratio, so the area of triangle $\triangle ABF$ is $4x$ . Now, consider trapezoid $ABED$ . Its area is $45+4x$ , which is three-fourths the area of the square. We set up an equation in $x$915\[45+4x = \frac{3}{4}\left(90+2x\right)\] Solving, we get $x = 9$ . The area of square $ABCD$ is $90+2x = 90 + 2 \cdot 9 = \boxed{108}$",108916267,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_22,2,"Point $E$ is the midpoint of side $\overline{CD}$ in square $ABCD,$ and $\overline{BE}$ meets diagonal $\overline{AC}$ at $F.$ The area of quadrilateral $AFED$ is $45.$ What is the area of $ABCD?$917[asy] size(5cm); draw((0,0)--(6,0)--(6,6)--(0,6)--cycle); draw((0,6)--(6,0)); draw((3,0)--(6,6)); label(""$A$"",(0,6),NW); label(""$B$"",(6,6),NE); label(""$C$"",(6,0),SE); label(""$D$"",(0,0),SW); label(""$E$"",(3,0),S); label(""$F$"",(4,2),E); [/asy]918$\textbf{(A) } 100 \qquad \textbf{(B) } 108 \qquad \textbf{(C) } 120 \qquad \textbf{(D) } 135 \qquad \textbf{(E) } 144$","We can use analytic geometry for this problem.919Let us start by giving $D$ the coordinate $(0,0)$ $A$ the coordinate $(0,1)$ , and so forth. $\overline{AC}$ and $\overline{EB}$ can be represented by the equations $y=-x+1$ and $y=2x-1$ , respectively. Solving for their intersection gives point $F$ coordinates $\left(\frac{2}{3},\frac{1}{3}\right)$920Now, $\triangle$ $EFC$ ’s area is simply $\frac{\frac{1}{2}\cdot\frac{1}{3}}{2}$ or $\frac{1}{12}$ . This means that pentagon $ABCEF$ ’s area is $\frac{1}{2}+\frac{1}{12}=\frac{7}{12}$ of the entire square, and it follows that quadrilateral $AFED$ ’s area is $\frac{5}{12}$ of the square.921The area of the square is then $\frac{45}{\frac{5}{12}}=9\cdot12=\boxed{108}$",108922268,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_22,3,"Point $E$ is the midpoint of side $\overline{CD}$ in square $ABCD,$ and $\overline{BE}$ meets diagonal $\overline{AC}$ at $F.$ The area of quadrilateral $AFED$ is $45.$ What is the area of $ABCD?$923[asy] size(5cm); draw((0,0)--(6,0)--(6,6)--(0,6)--cycle); draw((0,6)--(6,0)); draw((3,0)--(6,6)); label(""$A$"",(0,6),NW); label(""$B$"",(6,6),NE); label(""$C$"",(6,0),SE); label(""$D$"",(0,0),SW); label(""$E$"",(3,0),S); label(""$F$"",(4,2),E); [/asy]924$\textbf{(A) } 100 \qquad \textbf{(B) } 108 \qquad \textbf{(C) } 120 \qquad \textbf{(D) } 135 \qquad \textbf{(E) } 144$","$\triangle ABC$ has half the area of the square. $\triangle FEC$ has base equal to half the square side length, and by AA Similarity with $\triangle FBA$ , it has 1/(1+2)= 1/3 the height, so has $\dfrac1{12}$ th area of square. Thus, the area of the quadrilateral is $1-1/2-1/12=5/12$ th the area of the square. The area of the square is then $45\cdot\dfrac{12}{5}=\boxed{108}$",108925269,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_22,4,"Point $E$ is the midpoint of side $\overline{CD}$ in square $ABCD,$ and $\overline{BE}$ meets diagonal $\overline{AC}$ at $F.$ The area of quadrilateral $AFED$ is $45.$ What is the area of $ABCD?$926[asy] size(5cm); draw((0,0)--(6,0)--(6,6)--(0,6)--cycle); draw((0,6)--(6,0)); draw((3,0)--(6,6)); label(""$A$"",(0,6),NW); label(""$B$"",(6,6),NE); label(""$C$"",(6,0),SE); label(""$D$"",(0,0),SW); label(""$E$"",(3,0),S); label(""$F$"",(4,2),E); [/asy]927$\textbf{(A) } 100 \qquad \textbf{(B) } 108 \qquad \textbf{(C) } 120 \qquad \textbf{(D) } 135 \qquad \textbf{(E) } 144$","Extend $\overline{AD}$ and $\overline{BE}$ to meet at $X$ . Drop an altitude from $F$ to $\overline{CE}$ and call it $h$ . Also, call $\overline{CE}$ $x$ . As stated before, we have $\triangle ABF \sim \triangle CEF$ , so the ratio of their heights is in a $1:2$ ratio, making the altitude from $F$ to $\overline{AB}$ $2h$ . Note that this means that the side of the square is $3h$ . In addition, $\triangle XDE \sim \triangle XAB$ by AA Similarity in a $1:2$ ratio. This means that the side length of the square is $2x$ , making $3h=2x$928Now, note that $[ADEF]=[XAB]-[XDE]-[ABF]$ . We have $[\triangle XAB]=(4x)(2x)/2=4x^2,$ $[\triangle XDE]=(x)(2x)/2=x^2,$ and $[\triangle ABF]=(2x)(2h)/2=(2x)(4x/3)/2=4x^2/3.$ Subtracting makes $[ADEF]=4x^2-x^2-4x^2/3=5x^2/3.$ We are given that $[ADEF]=45,$ so $5x^2/3=45 \Rightarrow x^2=27.$ Therefore, $x= 3 \sqrt{3},$ so our answer is $(2x)^2=4x^2=4(27)=\boxed{108}.$",108929270,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_22,5,"Point $E$ is the midpoint of side $\overline{CD}$ in square $ABCD,$ and $\overline{BE}$ meets diagonal $\overline{AC}$ at $F.$ The area of quadrilateral $AFED$ is $45.$ What is the area of $ABCD?$930[asy] size(5cm); draw((0,0)--(6,0)--(6,6)--(0,6)--cycle); draw((0,6)--(6,0)); draw((3,0)--(6,6)); label(""$A$"",(0,6),NW); label(""$B$"",(6,6),NE); label(""$C$"",(6,0),SE); label(""$D$"",(0,0),SW); label(""$E$"",(3,0),S); label(""$F$"",(4,2),E); [/asy]931$\textbf{(A) } 100 \qquad \textbf{(B) } 108 \qquad \textbf{(C) } 120 \qquad \textbf{(D) } 135 \qquad \textbf{(E) } 144$","Solution with Cartesian and Barycentric Coordinates:932We start with the following:933Claim: Given a square $ABCD$ , let $E$ be the midpoint of $\overline{DC}$ and let $BE\cap AC = F$ . Then $\frac {AF}{FC}=2$934Proof: We use Cartesian coordinates. Let $D$ be the origin, $A=(0,1),C=(0,1),B=(1,1)$ . We have that $\overline{AC}$ and $\overline{EB}$ are governed by the equations $y=-x+1$ and $y=2x-1$ , respectively. Solving, $F=\left(\frac{2}{3},\frac{1}{3}\right)$ . The result follows. $\square$935Now, we apply Barycentric Coordinates w.r.t. $\triangle ACD$ . We let $A=(1,0,0),D=(0,1,0),C=(0,0,1)$ . Then $E=(0,\tfrac 12,\tfrac 12),F=(\tfrac 13,0,\tfrac 23)$936In the barycentric coordinate system, the area formula is $[XYZ]=\begin{vmatrix} x_{1} &y_{1}  &z_{1} \\ x_{2} &y_{2}  &z_{2} \\   x_{3}& y_{3} & z_{3} \end{vmatrix}\cdot [ABC]$ where $\triangle XYZ$ is a random triangle and $\triangle ABC$ is the reference triangle. Using this, we find that \[\frac{[FEC]}{[ACD]}=\begin{vmatrix} 0&0&1\\ 0&\tfrac 12&\tfrac 12\\ \tfrac 13&0&\tfrac 23 \end{vmatrix}=\frac16.\] Let $[FEC]=x$ so that $[ACD]=45+x$ . Then, we have $\frac{x}{x+45}=\frac 16 \Rightarrow x=9$ , so the answer is $2(45+9)=\boxed{108}$",108937271,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_23,1,"From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?938[asy] size(3cm); pair A[]; for (int i=0; i<9; ++i) {   A[i] = rotate(22.5+45*i)*(1,0); } filldraw(A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--A[6]--A[7]--cycle,gray,black); for (int i=0; i<8; ++i) { dot(A[i]); } [/asy]939$\textbf{(A) } \frac{2}{7} \qquad \textbf{(B) } \frac{5}{42} \qquad \textbf{(C) } \frac{11}{14} \qquad \textbf{(D) } \frac{5}{7} \qquad \textbf{(E) } \frac{6}{7}$","Choose side ""lengths"" $a,b,c$ for the triangle, where ""length"" is how many vertices of the octagon are skipped between vertices of the triangle, starting from the shortest side, and going clockwise, and choosing $a=b$ if the triangle is isosceles: $a+b+c=5$ , where either [ $a\leq b$ and $a < c$ ] or [ $a=b=c$ (but this is impossible in an octagon)].940Options are: $a=0$ with $b,c$ in { 0,5 ; 1,4 ; 2,3 ; 3,2 ; 4,1 }, and $a=1$ with { 1,3 ;  2,2} $5/7$ of these have a side with length 1, which corresponds to an edge of the octagon. So, our answer is $\boxed{57}$",57941272,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_23,2,"From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?942[asy] size(3cm); pair A[]; for (int i=0; i<9; ++i) {   A[i] = rotate(22.5+45*i)*(1,0); } filldraw(A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--A[6]--A[7]--cycle,gray,black); for (int i=0; i<8; ++i) { dot(A[i]); } [/asy]943$\textbf{(A) } \frac{2}{7} \qquad \textbf{(B) } \frac{5}{42} \qquad \textbf{(C) } \frac{11}{14} \qquad \textbf{(D) } \frac{5}{7} \qquad \textbf{(E) } \frac{6}{7}$","We will use constructive counting to solve this. There are $2$ cases: Either all $3$ points are adjacent, or exactly $2$ points are adjacent.944If all $3$ points are adjacent, then we have $8$ choices. If we have exactly $2$ adjacent points, then we will have $8$ places to put the adjacent points and $4$ places to put the remaining point, so we have $8\cdot4$ choices. The total amount of choices is ${8 \choose 3} = 8\cdot7$945Thus, our answer is $\frac{8+8\cdot4}{8\cdot7}= \frac{1+4}{7}=\boxed{57}$",57946273,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_23,3,"From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?947[asy] size(3cm); pair A[]; for (int i=0; i<9; ++i) {   A[i] = rotate(22.5+45*i)*(1,0); } filldraw(A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--A[6]--A[7]--cycle,gray,black); for (int i=0; i<8; ++i) { dot(A[i]); } [/asy]948$\textbf{(A) } \frac{2}{7} \qquad \textbf{(B) } \frac{5}{42} \qquad \textbf{(C) } \frac{11}{14} \qquad \textbf{(D) } \frac{5}{7} \qquad \textbf{(E) } \frac{6}{7}$","We can decide $2$ adjacent points with $8$ choices. The remaining point will have $6$ choices. However, we have counted the case with $3$ adjacent points twice, so we need to subtract this case once. The case with the $3$ adjacent points has $8$ arrangements, so our answer is949$\frac{8\cdot6-8}{{8 \choose 3 }}$ $=\frac{8\cdot6-8}{8 \cdot 7 \cdot 6 \div 6}\Longrightarrow\boxed{57}$",57950274,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_23,4,"From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?951[asy] size(3cm); pair A[]; for (int i=0; i<9; ++i) {   A[i] = rotate(22.5+45*i)*(1,0); } filldraw(A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--A[6]--A[7]--cycle,gray,black); for (int i=0; i<8; ++i) { dot(A[i]); } [/asy]952$\textbf{(A) } \frac{2}{7} \qquad \textbf{(B) } \frac{5}{42} \qquad \textbf{(C) } \frac{11}{14} \qquad \textbf{(D) } \frac{5}{7} \qquad \textbf{(E) } \frac{6}{7}$","Let $1$ point of the triangle be fixed at the top. Then, there are ${7 \choose 2} = 21$ ways to choose the other $2$ points. There must be $3$ spaces in the points and $3$ points themselves. This leaves $2$ extra points to be placed anywhere. By stars and bars, there are $3$ triangle points ( $n$ ) and $2$ extra points ( $k-1$ ) distributed so by the stars and bars formula, ${n+k-1 \choose k-1}$ , there are ${4 \choose 2} = 6$ ways to arrange the bars and stars. Thus, the probability is $\frac{(21 - 6)}{21} = \boxed{57}$",57953275,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_23,5,"From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?954[asy] size(3cm); pair A[]; for (int i=0; i<9; ++i) {   A[i] = rotate(22.5+45*i)*(1,0); } filldraw(A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--A[6]--A[7]--cycle,gray,black); for (int i=0; i<8; ++i) { dot(A[i]); } [/asy]955$\textbf{(A) } \frac{2}{7} \qquad \textbf{(B) } \frac{5}{42} \qquad \textbf{(C) } \frac{11}{14} \qquad \textbf{(D) } \frac{5}{7} \qquad \textbf{(E) } \frac{6}{7}$","We select a vertex of the octagon; this will be the first vertex of our triangle. Define the $distance$ of a vertex from another to be the minimum number of edges that one must travel on to get from one vertex to the other. There are three distinct cases; the second vertex is a distance of 1 away from the selected vertex (i.e. they are adjacent), the second vertex is a distance of 2 away from the selected vertex, or the second vertex is a distance of 3 or more away from the selected vertex. We consider each of these cases separately.956Case 1: The first two chosen vertices are adjacent.957There is a $\frac{2}{7}$ chance of selecting a point that is adjacent to the one we have chosen. In this case, any choice of the third vertex will result in a triangle that shares at least one side with the given octagon. Thus, this case has a $\frac{2}{7}$ case of giving us a triangle that fulfills the conditions given in the problem.958Case 2: There is a distance of 2 between the first two chosen vertices.959There is a $\frac{2}{7}$ chance of selecting a vertex that is a distance of 2 away from the first vertex. In this case, there are three vertices that will create a triangle that satisfies the condition in the problem (the one that is adjacent to both of the first two selected vertices and the two that are adjacent to only one of the first two selected vertices). There is a $\frac{1}{2}$ chance of selecting one of these three vertices from the remaining six. Thus, this case has a $\frac{2}{7} \cdot \frac{1}{2} = \frac{1}{7}$ chance of giving us a triangle that fulfills the conditions given in the problem.960Case 3: There is a distance of 3 or more between the first two chosen vertices.961There is a $\frac{3}{7}$ chance of selecting a vertex that is a distance of 3 or more away from the first vertex. In this case, there are four vertices (the two adjacent to each of the first two vertices) that may be chosen to satisfy the problem's condition. There is a $\frac{2}{3}$ chance of selecting one of these four vertices from the remaining six. Thus, this case has a $\frac{3}{7} \cdot \frac{2}{3} = \frac{2}{7}$ chance of giving us a triangle that fulfills the conditions given in the problem.962Summing the probabilities from each of the individual cases, we find that there is a $\frac{1}{7} + \frac{2}{7} + \frac{2}{7} = \boxed{57}$ chance of acquiring a triangle which shares at least one side with the octagon.",57963276,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_25,1,"How many perfect cubes lie between $2^8+1$ and $2^{18}+1$ , inclusive?964$\textbf{(A) }4\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }57\qquad \textbf{(E) }58$","We compute $2^8+1=257$ . We're all familiar with what $6^3$ is, namely $216$ , which is too small. The smallest cube greater than it is $7^3=343$ $2^{18}+1$ is too large to calculate, but we notice that $2^{18}=(2^6)^3=64^3$ , which therefore will clearly be the largest cube less than $2^{18}+1$ . So, the required number of cubes is $64-7+1= \boxed{58}$",58965277,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_25,2,"How many perfect cubes lie between $2^8+1$ and $2^{18}+1$ , inclusive?966$\textbf{(A) }4\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }57\qquad \textbf{(E) }58$","First, $2^8+1=257$ . Then, $2^{18}+1=262145$ . Now, we can see how many perfect cubes are between these two parameters. By guessing and checking, we find that it starts from $7$ and ends with $64$ . Now, by counting how many numbers are between these, we find the answer to be $\boxed{58}$",58967278,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_25,3,"How many perfect cubes lie between $2^8+1$ and $2^{18}+1$ , inclusive?968$\textbf{(A) }4\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }57\qquad \textbf{(E) }58$","First, we realize that question writers like to trick us. We know that most people will be calculating the lowest and highest number whose cubes are within the range. The answer will be the highest number $-$ the lowest number $+ 1$ . People will forget the $+1$ so the only possibilities are C and E. We can clearly see that C is too small so our answer is $\boxed{58}$",58969279,https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_25,4,"How many perfect cubes lie between $2^8+1$ and $2^{18}+1$ , inclusive?970$\textbf{(A) }4\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }57\qquad \textbf{(E) }58$","There is not so much guessing and checking after we find that it starts from $7$ because $7^3=343$ , which is over $2^8+1=257$ . We can start guessing with the 10, answer C, as it is the middle value. Adding 10 to 7 gives us 17 and $17^3 = 4,913$ , which is a bit low. So, we move ""up"" to 57, answer D. Adding 57 to 7 gives us 64 and $64^3 = 262,144$ , which is perfect for the $2^{18}+1=262,145$ . But we are not done. Since it is inclusive, we must add 1 to this solution, which give us $\boxed{58}$",58971280,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_2,1,"Alicia, Brenda, and Colby were the candidates in a recent election for student president. The pie chart below shows how the votes were distributed among the three candidates. If Brenda received $36$ votes, then how many votes were cast all together?972[asy] draw((-1,0)--(0,0)--(0,1)); draw((0,0)--(0.309, -0.951)); filldraw(arc((0,0), (0,1), (-1,0))--(0,0)--cycle, lightgray); filldraw(arc((0,0), (0.309, -0.951), (0,1))--(0,0)--cycle, gray); draw(arc((0,0), (-1,0), (0.309, -0.951))); label(""Colby"", (-0.5, 0.5)); label(""25\%"", (-0.5, 0.3)); label(""Alicia"", (0.7, 0.2)); label(""45\%"", (0.7, 0)); label(""Brenda"", (-0.5, -0.4)); label(""30\%"", (-0.5, -0.6)); [/asy]973$\textbf{(A) }70 \qquad \textbf{(B) }84 \qquad \textbf{(C) }100 \qquad \textbf{(D) }106 \qquad \textbf{(E) }120$","Let $x$ be the total amount of votes casted. From the chart, Brenda received $30\%$ of the votes and had $36$ votes. We can express this relationship as $\frac{30}{100}x=36$ . Solving for $x$ , we get $x=\boxed{120}.$",120974281,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_2,2,"Alicia, Brenda, and Colby were the candidates in a recent election for student president. The pie chart below shows how the votes were distributed among the three candidates. If Brenda received $36$ votes, then how many votes were cast all together?975[asy] draw((-1,0)--(0,0)--(0,1)); draw((0,0)--(0.309, -0.951)); filldraw(arc((0,0), (0,1), (-1,0))--(0,0)--cycle, lightgray); filldraw(arc((0,0), (0.309, -0.951), (0,1))--(0,0)--cycle, gray); draw(arc((0,0), (-1,0), (0.309, -0.951))); label(""Colby"", (-0.5, 0.5)); label(""25\%"", (-0.5, 0.3)); label(""Alicia"", (0.7, 0.2)); label(""45\%"", (0.7, 0)); label(""Brenda"", (-0.5, -0.4)); label(""30\%"", (-0.5, -0.6)); [/asy]976$\textbf{(A) }70 \qquad \textbf{(B) }84 \qquad \textbf{(C) }100 \qquad \textbf{(D) }106 \qquad \textbf{(E) }120$","We're being asked for the total number of votes cast -- that represents $100\%$ of the total number of votes. Brenda received $36$ votes, which is $\frac{30}{100} = \frac{3}{10}$ of the total number of votes. Multiplying $36$ by $\frac{10}{3},$ we get the total number of votes, which is $\boxed{120}.$",120977282,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_2,3,"Alicia, Brenda, and Colby were the candidates in a recent election for student president. The pie chart below shows how the votes were distributed among the three candidates. If Brenda received $36$ votes, then how many votes were cast all together?978[asy] draw((-1,0)--(0,0)--(0,1)); draw((0,0)--(0.309, -0.951)); filldraw(arc((0,0), (0,1), (-1,0))--(0,0)--cycle, lightgray); filldraw(arc((0,0), (0.309, -0.951), (0,1))--(0,0)--cycle, gray); draw(arc((0,0), (-1,0), (0.309, -0.951))); label(""Colby"", (-0.5, 0.5)); label(""25\%"", (-0.5, 0.3)); label(""Alicia"", (0.7, 0.2)); label(""45\%"", (0.7, 0)); label(""Brenda"", (-0.5, -0.4)); label(""30\%"", (-0.5, -0.6)); [/asy]979$\textbf{(A) }70 \qquad \textbf{(B) }84 \qquad \textbf{(C) }100 \qquad \textbf{(D) }106 \qquad \textbf{(E) }120$","If $36$ votes is $\frac{3}{10}$ of all the votes, we can divide that by $3$ to get $12$ as 10%, and then we can multiply the $12$ by $10$ to get to $120$ . So, the answer is $\boxed{120}.$",120980283,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_3,1,"What is the value of the expression $\sqrt{16\sqrt{8\sqrt{4}}}$981$\textbf{(A) }4\qquad\textbf{(B) }4\sqrt{2}\qquad\textbf{(C) }8\qquad\textbf{(D) }8\sqrt{2}\qquad\textbf{(E) }16$",$\sqrt{16\sqrt{8\sqrt{4}}}$ $\sqrt{16\sqrt{8\cdot 2}}$ $\sqrt{16\sqrt{16}}$ $\sqrt{16\cdot 4}$ $\sqrt{64}$ $\boxed{8}$,8982284,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_4,1,"When $0.000315$ is multiplied by $7,928,564$ the product is closest to which of the following?983$\textbf{(A) }210\qquad\textbf{(B) }240\qquad\textbf{(C) }2100\qquad\textbf{(D) }2400\qquad\textbf{(E) }24000$","We can approximate $7,928,564$ to $8,000,000$ and $0.000315$ to $0.0003.$ Multiplying the two yields $2400.$ Thus, it shows our answer is $\boxed{2400}.$",400984285,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_5,1,"What is the value of the expression $\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}$985$\textbf{(A) }1020\qquad\textbf{(B) }1120\qquad\textbf{(C) }1220\qquad\textbf{(D) }2240\qquad\textbf{(E) }3360$","Directly calculating:986We evaluate both the top and bottom: $\frac{40320}{36}$ . This simplifies to $\boxed{1120}$",120987286,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_5,2,"What is the value of the expression $\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}$988$\textbf{(A) }1020\qquad\textbf{(B) }1120\qquad\textbf{(C) }1220\qquad\textbf{(D) }2240\qquad\textbf{(E) }3360$","It is well known that the sum of all numbers from $1$ to $n$ is $\frac{n(n+1)}{2}$ . Therefore, the denominator is equal to $\frac{8 \cdot 9}{2} = 4 \cdot 9 = 2 \cdot 3 \cdot 6$ . Now, we can cancel the factors of $2$ $3$ , and $6$ from both the numerator and denominator, only leaving $8 \cdot 7 \cdot 5 \cdot 4 \cdot 1$ . This evaluates to $\boxed{1120}$",120989287,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_5,3,"What is the value of the expression $\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}$990$\textbf{(A) }1020\qquad\textbf{(B) }1120\qquad\textbf{(C) }1220\qquad\textbf{(D) }2240\qquad\textbf{(E) }3360$","First, we evaluate $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8$ to get 36. We notice that 36 is 6 squared, so we can factor the denominator like $\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{6 \cdot 6}$ then cancel the 6s out,\ to get $\frac{4 \cdot 5 \cdot 7 \cdot 8}{1}$ . Now that we have escaped fraction form, we multiply $4 \cdot 5 \cdot 7 \cdot 8$ . Multiplying these, we get $\boxed{1120}$",120991288,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_6,1,"If the degree measures of the angles of a triangle are in the ratio $3:3:4$ , what is the degree measure of the largest angle of the triangle?992$\textbf{(A) }18\qquad\textbf{(B) }36\qquad\textbf{(C) }60\qquad\textbf{(D) }72\qquad\textbf{(E) }90$","The sum of the ratios is $10$ .  Since the sum of the angles of a triangle is $180^{\circ}$ , the ratio can be scaled up to $54:54:72$ $(3\cdot 18:3\cdot 18:4\cdot 18).$ The numbers in the ratio $54:54:72$ represent the angles of the triangle.  The question asks for the largest, so the answer is $\boxed{72}$",72993289,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_6,2,"If the degree measures of the angles of a triangle are in the ratio $3:3:4$ , what is the degree measure of the largest angle of the triangle?994$\textbf{(A) }18\qquad\textbf{(B) }36\qquad\textbf{(C) }60\qquad\textbf{(D) }72\qquad\textbf{(E) }90$","We can denote the angles of the triangle as $3x$ $3x$ $4x$ . Due to the sum of the angles in a triangle, $3x+3x+4x=180^{\circ}\implies x=18^{\circ}$ . The greatest angle is $4x$ and after substitution we get $\boxed{72}$",72995290,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_6,3,"If the degree measures of the angles of a triangle are in the ratio $3:3:4$ , what is the degree measure of the largest angle of the triangle?996$\textbf{(A) }18\qquad\textbf{(B) }36\qquad\textbf{(C) }60\qquad\textbf{(D) }72\qquad\textbf{(E) }90$","We know the longest side must be denoted by the 4 in the ratio. Since the ratio is 3:3:4, we know that the longest side must be $\frac{4}{3+3+4}$ of the degree total (which for all triangles is 180). Thus, \[\frac{4}{3+3+4} \cdot 180 = \frac{4}{10} \cdot 180 = \boxed{72}\]",72997291,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_6,4,"If the degree measures of the angles of a triangle are in the ratio $3:3:4$ , what is the degree measure of the largest angle of the triangle?998$\textbf{(A) }18\qquad\textbf{(B) }36\qquad\textbf{(C) }60\qquad\textbf{(D) }72\qquad\textbf{(E) }90$","Since we see the ratio is $3:3:4$ , we can rule out the answer of ${\textbf{(E) }90}$ because the numbers in the ratio are too big to have $90^\circ$ . Also, we are trying to find the largest angle and all the other angles except for 72 are too small to be the largest angle. Using all this, our answer is $\boxed{72}$",72999292,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_7,1,"Let $Z$ be a 6-digit positive integer, such as 247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of $Z$1000$\textbf{(A) }11\qquad\textbf{(B) }19\qquad\textbf{(C) }101\qquad\textbf{(D) }111\qquad\textbf{(E) }1111$","To check, if a number is divisible by 19, take its unit digit and multiply it by 2, then add the result to the rest of the number, and repeat this step until the number is reduced to two digits. If the result is divisible by 19, then the original number is also divisible by 19. Or we could just try to divide the example number the problem gave us by 19.1001After we got 19 eliminated, we can see that the other options have a lot of 1's in them. The divisibility rule for 11 is add alternating digits up, then take the difference of them. The example number works like that. If we add variables, ABCDEF to make number ABCABC, we can see that (A+C+B) - (B+A+C) = 0. Which is divisible by 11, so our answer choice is $\boxed{11}$",111002293,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_7,2,"Let $Z$ be a 6-digit positive integer, such as 247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of $Z$1003$\textbf{(A) }11\qquad\textbf{(B) }19\qquad\textbf{(C) }101\qquad\textbf{(D) }111\qquad\textbf{(E) }1111$","We are given one of the numbers that can represent $Z$ , so we can just try out the options to see which one is a factor of $247247$ . We get $\boxed{11}$",111004294,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_7,3,"Let $Z$ be a 6-digit positive integer, such as 247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of $Z$1005$\textbf{(A) }11\qquad\textbf{(B) }19\qquad\textbf{(C) }101\qquad\textbf{(D) }111\qquad\textbf{(E) }1111$","To find out when a number is divisible by 11, place plus and minus signs alternatively in front of every digit, then calculate the result. If this result is divisible by 11 (including 0), the number is divisible by 11; otherwise, the number isn’t divisible by 11. In this case, $+2-4+7-2+4-7=0$ . Because the result is 0, the number 247247 is divisible by 11  and so we get $\boxed{11}$",111006295,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_7,4,"Let $Z$ be a 6-digit positive integer, such as 247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of $Z$1007$\textbf{(A) }11\qquad\textbf{(B) }19\qquad\textbf{(C) }101\qquad\textbf{(D) }111\qquad\textbf{(E) }1111$","Similar to solution 1, let $Z=ABCABC$ . To prove it is divisible by 11, we can compute its alternating sum, which is $A-B+C-A+B-C=0$ , which is divisible by 11. Therefore, the answer is $\boxed{11}$",111008296,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_7,5,"Let $Z$ be a 6-digit positive integer, such as 247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of $Z$1009$\textbf{(A) }11\qquad\textbf{(B) }19\qquad\textbf{(C) }101\qquad\textbf{(D) }111\qquad\textbf{(E) }1111$","We can find that all numbers like $Z$ are divisible by 1001. 1001 is divisible by 11 because when we divide it, we get a whole number. So, the answer is $\boxed{11}$",111010297,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_8,1,"Malcolm wants to visit Isabella after school today and knows the street where she lives but doesn't know her house number. She tells him, ""My house number has two digits, and exactly three of the following four statements about it are true.""1011(1) It is prime.1012(2) It is even.1013(3) It is divisible by 7.1014(4) One of its digits is 9.1015This information allows Malcolm to determine Isabella's house number. What is its units digit?1016$\textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }8\qquad\textbf{(E) }9$","Notice that (1) cannot be true. Otherwise, the number would have to be prime and be either even or divisible by 7. This only happens if the number is 2 or 7, neither of which are two-digit numbers, so we run into a contradiction. Thus, we must have (2), (3), and (4) be true. By (2), the $2$ -digit number is even, and thus, the digit in the tens place must be $9$ . The only even $2$ -digit number starting with $9$ and divisible by $7$ is $98$ , which has a units digit of $\boxed{8}.$",81017298,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_8,2,"Malcolm wants to visit Isabella after school today and knows the street where she lives but doesn't know her house number. She tells him, ""My house number has two digits, and exactly three of the following four statements about it are true.""1018(1) It is prime.1019(2) It is even.1020(3) It is divisible by 7.1021(4) One of its digits is 9.1022This information allows Malcolm to determine Isabella's house number. What is its units digit?1023$\textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }8\qquad\textbf{(E) }9$","(Statement 1) Cannot be true, because only one of these four statements is true, and (Statement 1) states that the number is prime, which would make (Statement 2) and (Statement 3) false, which is not possible. And since the number being described is even, it must end with an even number (0,2,4,6,8). And since the number being described is a two-digit number, the first digit must be 9 (according to statement 4) because we are looking for the units digit (the digit in the one's place of a number). And so if we plug in the number 9 to all of the even answers, we will get three possible outcomes. $94$ $96$ , and $98$ . Because the number described is divisible by 7 (according to statement 3), the only possible answer for the number being described would be $98$ . So, the answer is $\boxed{8}.$",81024299,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_8,3,"Malcolm wants to visit Isabella after school today and knows the street where she lives but doesn't know her house number. She tells him, ""My house number has two digits, and exactly three of the following four statements about it are true.""1025(1) It is prime.1026(2) It is even.1027(3) It is divisible by 7.1028(4) One of its digits is 9.1029This information allows Malcolm to determine Isabella's house number. What is its units digit?1030$\textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }8\qquad\textbf{(E) }9$","Like solutions 1 and 2, Statement 1 can't be true because it would contradict both Statements 2 and 3. Therefore, the other three must be true. We know the following:1031The only multiple of 14 that has a tens digit of 9 is 98. Thus, our answer is $\boxed{8}.$",81032300,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_9,1,"All of Marcy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the smallest number of yellow marbles that Macy could have?1033$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }5$","The $6$ green marbles and yellow marbles form $1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}$ of the total marbles. Now, suppose the total number of marbles is $x$ . We know the number of yellow marbles is $\frac{5}{12}x - 6$ and a positive integer. Therefore, $12$ must divide $x$ . Trying the smallest multiples of $12$ for $x$ , we see that when $x = 12$ , we get there are $-1$ yellow marbles, which is impossible. However when $x = 24$ , there are $\frac{5}{12} \cdot 24 - 6 = \boxed{4}$ yellow marbles, which must be the smallest possible.",41034301,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_9,2,"All of Marcy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the smallest number of yellow marbles that Macy could have?1035$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }5$","Since $\frac{1}{3}$ of the marbles are blue and $\frac{1}{4}$ are red, it is clear that the total number of marbles must be divisible by $12$ . If there are $12$ marbles, then $4$ are blue, $3$ are red, and $6$ are green, meaning that there are $-1$ yellow marbles. This is impossible. Trying the next multiple of $12$ $24$ , we find that $8$ are green, $6$ are red, and $6$ are green, meaning that the minimum number of yellow marbles is $\boxed{4}$",41036302,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_11,1,"A square-shaped floor is covered with congruent square tiles. If the total number of tiles that lie on the two diagonals is 37, how many tiles cover the floor?1037$\textbf{(A) }148\qquad\textbf{(B) }324\qquad\textbf{(C) }361\qquad\textbf{(D) }1296\qquad\textbf{(E) }1369$","Since the number of tiles lying on both diagonals is $37$ , counting one tile twice, there are $37=2x-1\implies x=19$ tiles on each side. Therefore, our answer is $19^2=361=\boxed{361}$",3611038303,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_11,2,"A square-shaped floor is covered with congruent square tiles. If the total number of tiles that lie on the two diagonals is 37, how many tiles cover the floor?1039$\textbf{(A) }148\qquad\textbf{(B) }324\qquad\textbf{(C) }361\qquad\textbf{(D) }1296\qquad\textbf{(E) }1369$","Visualize it as 4 separate diagonals connecting to one square in the middle. Each square on the diagonal corresponds to one square of horizontal/vertical distance (because it's a square). So, we figure out the length of each separate diagonal, multiply by two, and then add 1. (Realize that we can just join two of the separate diagonals on opposite sides together to save some time in calculations.) Therefore, the edge length is: \[\frac{37-1}{4} \cdot 2 + 1 = 19\] Thus, our solution is $19^2 = 361 = \boxed{361}$",3611040304,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_14,1,"Chloe and Zoe are both students in Ms. Demeanor's math class. Last night, they each solved half of the problems in their homework assignment alone and then solved the other half together. Chloe had correct answers to only $80\%$ of the problems she solved alone, but overall $88\%$ of her answers were correct. Zoe had correct answers to $90\%$ of the problems she solved alone. What was Zoe's  overall percentage of correct answers?1041$\textbf{(A) }89\qquad\textbf{(B) }92\qquad\textbf{(C) }93\qquad\textbf{(D) }96\qquad\textbf{(E) }98$","Let the number of questions that they solved alone be $x$ . Let the percentage of problems they correctly solve together be $a$ %. 1042As given, \[\frac{80x}{100} + \frac{ax}{100} = \frac{2 \cdot 88x}{100}\]1043Hence, $a = 96$1044Zoe got $\frac{90x}{100} + \frac{ax}{100} = \frac{186x}{100}$ problems right out of $2x$ . Therefore, Zoe got $\frac{\frac{186x}{100}}{2x} = \frac{93}{100} = \boxed{93}$ percent of the problems correct.",931045305,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_14,2,"Chloe and Zoe are both students in Ms. Demeanor's math class. Last night, they each solved half of the problems in their homework assignment alone and then solved the other half together. Chloe had correct answers to only $80\%$ of the problems she solved alone, but overall $88\%$ of her answers were correct. Zoe had correct answers to $90\%$ of the problems she solved alone. What was Zoe's  overall percentage of correct answers?1046$\textbf{(A) }89\qquad\textbf{(B) }92\qquad\textbf{(C) }93\qquad\textbf{(D) }96\qquad\textbf{(E) }98$","Assume the total amount of problems is $100$ per half homework assignment since we are dealing with percentages, not values. Then, we know that Chloe got $80$ problems correct by herself and got $176$ problems correct overall. We also know that Zoe had $90$ problems she did correctly alone. We can see that the total amount of correct problems Chloe and Zoe did together was $176-80=96$ . Therefore, Zoe did $96+90=186$ problems out of $200$ problems correctly. This is $\boxed{93}$ percent.",931047306,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_14,3,"Chloe and Zoe are both students in Ms. Demeanor's math class. Last night, they each solved half of the problems in their homework assignment alone and then solved the other half together. Chloe had correct answers to only $80\%$ of the problems she solved alone, but overall $88\%$ of her answers were correct. Zoe had correct answers to $90\%$ of the problems she solved alone. What was Zoe's  overall percentage of correct answers?1048$\textbf{(A) }89\qquad\textbf{(B) }92\qquad\textbf{(C) }93\qquad\textbf{(D) }96\qquad\textbf{(E) }98$","In the problem, we can see that Chloe solved 80% of the problems she solved alone, but 88% of her answers are correct. If 80 and another number's average is 88, the other number must be 96. Then, Zoe solved 90% of the problems she did alone, but 96% of her answers were correct. Then, the average of 90 and 96 is $\boxed{93}$",931049307,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_14,4,"Chloe and Zoe are both students in Ms. Demeanor's math class. Last night, they each solved half of the problems in their homework assignment alone and then solved the other half together. Chloe had correct answers to only $80\%$ of the problems she solved alone, but overall $88\%$ of her answers were correct. Zoe had correct answers to $90\%$ of the problems she solved alone. What was Zoe's  overall percentage of correct answers?1050$\textbf{(A) }89\qquad\textbf{(B) }92\qquad\textbf{(C) }93\qquad\textbf{(D) }96\qquad\textbf{(E) }98$","(Slightly different Solution)1051Suppose we said that there were $100$ problems in their assignment. Then, Chloe had $40$ correct and $10$ incorrect on her portion, and $48$ correct and $2$ incorrect on the portion she and Zoe solved. Zoe has $45$ correct and $5$ incorrect on her portion, and $48$ correct and $2$ incorrect on the portion that she and Chloe solved. Then, Zoe has $48 + 45 = 93$ correct answers out of $100$ , so the answer is $\boxed{93}$",931052308,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_14,5,"Chloe and Zoe are both students in Ms. Demeanor's math class. Last night, they each solved half of the problems in their homework assignment alone and then solved the other half together. Chloe had correct answers to only $80\%$ of the problems she solved alone, but overall $88\%$ of her answers were correct. Zoe had correct answers to $90\%$ of the problems she solved alone. What was Zoe's  overall percentage of correct answers?1053$\textbf{(A) }89\qquad\textbf{(B) }92\qquad\textbf{(C) }93\qquad\textbf{(D) }96\qquad\textbf{(E) }98$","Let the total number of problems be $t$ . Let the percentage of the number of problems that Chloe and Zoe did together and got right be $x$ . As we can see, Chloe got $80$ % of $\frac {1}{2}$ of the total problems right, hence, ${0.80 \cdot \frac{1}{2}t}$ . We also know that Chloe got $88$ % of $t$ problems right altogether, making it ${0.88 \cdot t}$ total problems right. If we add $x$ to the percentage of correct problems that Chloe solved alone, then that should be equal to the total number of problems that Chloe got right, making the equation: ${0.80 \cdot \frac{1}{2}t} + x = {0.88 \cdot t}$ . Solving that, we get $x = 0.48t$ . We also know that Zoe got $90$ % of $\frac {1}{2}$ of the total problems right, making it ${0.90 \cdot \frac{1}{2}t}$ . We now add that amount to the percentage of problems that Chloe and Zoe got right together, making ${0.90 \cdot \frac{1}{2}t}+ 0.48t$ . Solving that, we get $0.93t$ , which is equal to $93$ %; hence, $\boxed{93}$",931054309,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_15,1,"In the arrangement of letters and numerals below, by how many different paths can one spell AMC8? Beginning at the A in the middle, a path only allows moves from one letter to an adjacent (above, below, left, or right, but not diagonal) letter. One example of such a path is traced in the picture.1055[asy] fill((0.5, 4.5)--(1.5,4.5)--(1.5,2.5)--(0.5,2.5)--cycle,lightgray); fill((1.5,3.5)--(2.5,3.5)--(2.5,1.5)--(1.5,1.5)--cycle,lightgray); label(""$8$"", (1, 0)); label(""$C$"", (2, 0)); label(""$8$"", (3, 0)); label(""$8$"", (0, 1)); label(""$C$"", (1, 1)); label(""$M$"", (2, 1)); label(""$C$"", (3, 1)); label(""$8$"", (4, 1)); label(""$C$"", (0, 2)); label(""$M$"", (1, 2)); label(""$A$"", (2, 2)); label(""$M$"", (3, 2)); label(""$C$"", (4, 2)); label(""$8$"", (0, 3)); label(""$C$"", (1, 3)); label(""$M$"", (2, 3)); label(""$C$"", (3, 3)); label(""$8$"", (4, 3)); label(""$8$"", (1, 4)); label(""$C$"", (2, 4)); label(""$8$"", (3, 4));[/asy]1056$\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }12\qquad\textbf{(D) }24\qquad\textbf{(E) }36$","Notice that the upper-most section contains a 3 by 3 square that looks like:1057[asy]label(""$8$"", (1, 2)); label(""$C$"", (2, 2)); label(""$8$"", (3, 2)); label(""$C$"", (1, 1)); label(""$M$"", (2, 1)); label(""$C$"", (3, 1)); label(""$M$"", (1, 0)); label(""$A$"", (2, 0)); label(""$M$"", (3, 0));[/asy]1058It has 6 paths in which you can spell out AMC8. You will find four identical copies of this square in the figure, so multiply ${6 \cdot 4}$ to get $\boxed{24}$ total paths.",241059310,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_15,2,"In the arrangement of letters and numerals below, by how many different paths can one spell AMC8? Beginning at the A in the middle, a path only allows moves from one letter to an adjacent (above, below, left, or right, but not diagonal) letter. One example of such a path is traced in the picture.1060[asy] fill((0.5, 4.5)--(1.5,4.5)--(1.5,2.5)--(0.5,2.5)--cycle,lightgray); fill((1.5,3.5)--(2.5,3.5)--(2.5,1.5)--(1.5,1.5)--cycle,lightgray); label(""$8$"", (1, 0)); label(""$C$"", (2, 0)); label(""$8$"", (3, 0)); label(""$8$"", (0, 1)); label(""$C$"", (1, 1)); label(""$M$"", (2, 1)); label(""$C$"", (3, 1)); label(""$8$"", (4, 1)); label(""$C$"", (0, 2)); label(""$M$"", (1, 2)); label(""$A$"", (2, 2)); label(""$M$"", (3, 2)); label(""$C$"", (4, 2)); label(""$8$"", (0, 3)); label(""$C$"", (1, 3)); label(""$M$"", (2, 3)); label(""$C$"", (3, 3)); label(""$8$"", (4, 3)); label(""$8$"", (1, 4)); label(""$C$"", (2, 4)); label(""$8$"", (3, 4));[/asy]1061$\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }12\qquad\textbf{(D) }24\qquad\textbf{(E) }36$","There are three different kinds of paths that are on this diagram. The first kind is when you directly count $A$ $M$ $C$ in a straight line. The second is when you count $A$ , turn left or right to get $M$ , then go up or down to count $8$ and $C$ . The third is the one where you start with $A$ , move up or down to count $M$ , turn left or right to count $C$ , then move straight again to get $8$1062There are 8 paths for each kind of path, making for $8 \cdot 3=\boxed{24}$ paths.",241063311,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_15,3,"In the arrangement of letters and numerals below, by how many different paths can one spell AMC8? Beginning at the A in the middle, a path only allows moves from one letter to an adjacent (above, below, left, or right, but not diagonal) letter. One example of such a path is traced in the picture.1064[asy] fill((0.5, 4.5)--(1.5,4.5)--(1.5,2.5)--(0.5,2.5)--cycle,lightgray); fill((1.5,3.5)--(2.5,3.5)--(2.5,1.5)--(1.5,1.5)--cycle,lightgray); label(""$8$"", (1, 0)); label(""$C$"", (2, 0)); label(""$8$"", (3, 0)); label(""$8$"", (0, 1)); label(""$C$"", (1, 1)); label(""$M$"", (2, 1)); label(""$C$"", (3, 1)); label(""$8$"", (4, 1)); label(""$C$"", (0, 2)); label(""$M$"", (1, 2)); label(""$A$"", (2, 2)); label(""$M$"", (3, 2)); label(""$C$"", (4, 2)); label(""$8$"", (0, 3)); label(""$C$"", (1, 3)); label(""$M$"", (2, 3)); label(""$C$"", (3, 3)); label(""$8$"", (4, 3)); label(""$8$"", (1, 4)); label(""$C$"", (2, 4)); label(""$8$"", (3, 4));[/asy]1065$\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }12\qquad\textbf{(D) }24\qquad\textbf{(E) }36$","Notice that the $A$ is adjacent to $4$ $M$ s, each $M$ is adjacent to $3$ $C$ s, and each $C$ is adjacent to $2$ $8$ 's. So for each $A$ , there are $4$ $M$ s, and for each $M$ , there are $3$ $C$ s, and for each $C$ , there are $2$ $8$ s. Thus, the answer is $1\cdot 4\cdot 3\cdot 2 = \boxed{24}.$",241066312,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_15,4,"In the arrangement of letters and numerals below, by how many different paths can one spell AMC8? Beginning at the A in the middle, a path only allows moves from one letter to an adjacent (above, below, left, or right, but not diagonal) letter. One example of such a path is traced in the picture.1067[asy] fill((0.5, 4.5)--(1.5,4.5)--(1.5,2.5)--(0.5,2.5)--cycle,lightgray); fill((1.5,3.5)--(2.5,3.5)--(2.5,1.5)--(1.5,1.5)--cycle,lightgray); label(""$8$"", (1, 0)); label(""$C$"", (2, 0)); label(""$8$"", (3, 0)); label(""$8$"", (0, 1)); label(""$C$"", (1, 1)); label(""$M$"", (2, 1)); label(""$C$"", (3, 1)); label(""$8$"", (4, 1)); label(""$C$"", (0, 2)); label(""$M$"", (1, 2)); label(""$A$"", (2, 2)); label(""$M$"", (3, 2)); label(""$C$"", (4, 2)); label(""$8$"", (0, 3)); label(""$C$"", (1, 3)); label(""$M$"", (2, 3)); label(""$C$"", (3, 3)); label(""$8$"", (4, 3)); label(""$8$"", (1, 4)); label(""$C$"", (2, 4)); label(""$8$"", (3, 4));[/asy]1068$\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }12\qquad\textbf{(D) }24\qquad\textbf{(E) }36$","We can do this problem by computing how many ways there are to get to each letter (in order). There is $1$ way to get to the $A$ in the center. We can only get to each of the other $M$ s by going there from the $A$ , so there is $1$ way to get to each of the four $M$ s. For the $C$ s, we notice that four $C$ s are surrounded by one $M$ , and four $C$ s are surrounded by two $M$ s. Finally, each of the $8$ s is surrounded by one $C$ with one way to get there and one $C$ with two ways to get there. Therefore, there are three paths to any of the $8$ s. Since there are eight $8$ s, the answer is $3 \cdot 8 = \boxed{24}.$",241069313,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_17,2,"Starting with some gold coins and some empty treasure chests, I tried to put $9$ gold coins in each treasure chest, but that left $2$ treasure chests empty.  So instead I put $6$ gold coins in each treasure chest, but then I had $3$ gold coins left over.  How many gold coins did I have?1070$\textbf{(A) }9\qquad\textbf{(B) }27\qquad\textbf{(C) }45\qquad\textbf{(D) }63\qquad\textbf{(E) }81$","With $9$ coins, there are $\frac{9}{9}+2=1+2=3$ chests, by the first condition. These don't fit in with the second condition, so we move onto $27$ coins. By the same first condition, there are $5$ chests( $\frac{27}{9}+2$ ). This also doesn't fit with the second condition. So, onto $45$ coins. The first condition implies that there are $\frac{45}{9}+2=7$ chests, which DOES fit with the second condition, since $6\cdot7+3=42+3=45$ . Thus, the desired value is $\boxed{45}$",451071314,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_18,1,"In the non-convex quadrilateral $ABCD$ shown below, $\angle BCD$ is a right angle, $AB=12$ $BC=4$ $CD=3$ , and $AD=13$ . What is the area of quadrilateral $ABCD$1072[asy]draw((0,0)--(2.4,3.6)--(0,5)--(12,0)--(0,0)); label(""$B$"", (0, 0), SW); label(""$A$"", (12, 0), ESE); label(""$C$"", (2.4, 3.6), SE); label(""$D$"", (0, 5), N);[/asy]1073$\textbf{(A) }12 \qquad \textbf{(B) }24 \qquad \textbf{(C) }26 \qquad \textbf{(D) }30 \qquad \textbf{(E) }36$","We first connect point $B$ with point $D$1074[asy]draw((0,0)--(2.4,3.6)--(0,5)--(12,0)--(0,0)); draw((0,0)--(0,5)); label(""$B$"", (0, 0), SW); label(""$A$"", (12, 0), ESE); label(""$C$"", (2.4, 3.6), SE); label(""$D$"", (0, 5), N);[/asy]1075We can see that $\triangle BCD$ is a 3-4-5 right triangle. We can also see that $\triangle BDA$ is a right triangle, by the 5-12-13 Pythagorean triple. With these lengths, we can solve the problem. The area of $\triangle BDA$ is $\frac{5\cdot 12}{2}$ , and the area of $\triangle BCD$ is $\frac{3\cdot 4}{2}$ . Thus, the area of quadrilateral $ABCD$ is $30-6 = \boxed{24}.$",241076315,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_18,2,"In the non-convex quadrilateral $ABCD$ shown below, $\angle BCD$ is a right angle, $AB=12$ $BC=4$ $CD=3$ , and $AD=13$ . What is the area of quadrilateral $ABCD$1077[asy]draw((0,0)--(2.4,3.6)--(0,5)--(12,0)--(0,0)); label(""$B$"", (0, 0), SW); label(""$A$"", (12, 0), ESE); label(""$C$"", (2.4, 3.6), SE); label(""$D$"", (0, 5), N);[/asy]1078$\textbf{(A) }12 \qquad \textbf{(B) }24 \qquad \textbf{(C) }26 \qquad \textbf{(D) }30 \qquad \textbf{(E) }36$","$\triangle BCD$ is a 3-4-5 right triangle. So the area of $\triangle BCD$ is 6. Then we can use Heron's formula to compute the area of $\triangle ABD$ whose sides have lengths 5,12,and 13. The area of $\triangle ABD$ $\sqrt{s(s-5)(s-12)(s-13)}$ , where s is the semi-perimeter of the triangle, that is $s=(5+12+13)/2=15.$ Thus, the area of $\triangle ABD$ is 30, so the area of $ABCD$ is $30-6 = \boxed{24}.$ ---LarryFlora",241079316,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_19,2,"For any positive integer $M$ , the notation $M!$ denotes the product of the integers $1$ through $M$ . What is the largest integer $n$ for which $5^n$ is a factor of the sum $98!+99!+100!$1080$\textbf{(A) }23\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$","Also, keep in mind that the number of $5$ ’s in $98! (10,000)$ is the same as the number of trailing zeros. The number of zeros is $98!$ , which means we need pairs of $5$ ’s and $2$ ’s; we know there will be many more $2$ ’s, so we seek to find the number of $5$ ’s in $98!$ , which the solution tells us. And, that is $22$ factors of $5$ $10,000$ has $4$ trailing zeros, so it has $4$ factors of $5$ and $22 + 4 = \boxed{26}$",261081317,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_19,3,"For any positive integer $M$ , the notation $M!$ denotes the product of the integers $1$ through $M$ . What is the largest integer $n$ for which $5^n$ is a factor of the sum $98!+99!+100!$1082$\textbf{(A) }23\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$","We can first factor a $98!$ out of the $98! + 99! + 100!$ to get $98! ( 1 + 99 + 100*99 ),$ Simplify to get $98! (10,000)$1083Let's first find how many factors of $5 10,000$ has. $10,000$ is $(2*5)^4$ because $10,000$ is $(10)^4$ . After we remove the brackets, we get $2^4$ , and $5^4$ . We only care about the latter (second one), because the problem only ask's for the power of $5$ . We get $4$1084Next, we can look at the multiples of 5 in $98!$ $98/5 = 19$ so there is 19 multiples of 5. We get $19$1085But we cannot forget the multiples of $5$ with $2$ fives in it. Multiples of $25$ . How many multiples of $25$ are between $1$ and $98$ $3$ $25,50,75,$ and that's it. We get $3$1086Finally, we add all of the numbers (powers of $5$ ) up. That is $4 + 19 + 3$ , which is just $26$1087So the answer is $26$ . Which is answer choice D $\boxed{26}$",261088318,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_21,1,"Suppose $a$ $b$ , and $c$ are nonzero real numbers, and $a+b+c=0$ . What are the possible value(s) for $\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}$1089$\textbf{(A) }0\qquad\textbf{(B) }1\text{ and }-1\qquad\textbf{(C) }2\text{ and }-2\qquad\textbf{(D) }0,2,\text{ and }-2\qquad\textbf{(E) }0,1,\text{ and }-1$","There are $2$ cases to consider:1090Case $1$ $2$ of $a$ $b$ , and $c$ are positive and the other is negative. Without loss of generality (WLOG), we can assume that $a$ and $b$ are positive and $c$ is negative. In this case, we have that \[\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}=1+1-1-1=0.\]1091Case $2$ $2$ of $a$ $b$ , and $c$ are negative and the other is positive. WLOG, we can assume that $a$ and $b$ are negative and $c$ is positive. In this case, we have that \[\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}=-1-1+1+1=0.\]1092Note these are the only valid cases, for neither $3$ negatives nor $3$ positives would work as they cannot sum up to $0$ . In both cases, we get that the given expression equals $\boxed{0}$",01093319,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_23,1,"Each day for four days, Linda traveled for one hour at a speed that resulted in her traveling one mile in an integer number of minutes. Each day after the first, her speed decreased so that the number of minutes to travel one mile increased by $5$ minutes over the preceding day. Each of the four days, her distance traveled was also an integer number of miles. What was the total number of miles for the four trips?1094$\textbf{(A) }10\qquad\textbf{(B) }15\qquad\textbf{(C) }25\qquad\textbf{(D) }50\qquad\textbf{(E) }82$","It is well known that $\text{Distance}=\text{Speed} \cdot \text{Time}$ . In the question, we want distance. From the question, we have that the time is $60$ minutes or $1$ hour.  By the equation derived from $\text{Distance}=\text{Speed} \cdot \text{Time}$ , we have $\text{Speed}=\frac{\text{Distance}}{\text{Time}}$ , so the speed is $1$ mile per $x$ minutes. Because we want the distance, we multiply the time and speed together yielding $60\text{ mins}\cdot \frac{1\text{ mile}}{x\text{ mins}}$ . The minutes cancel out, so now we have $\dfrac{60}{x}$ as our distance for the first day. The distance for the following days are: \[\dfrac{60}{x},\dfrac{60}{x+5},\dfrac{60}{x+10},\dfrac{60}{x+15}.\] We know that $x,x+5,x+10,x+15$ are all factors of $60$ , therefore, $x=5$ because the factors have to be in an arithmetic sequence with the common difference being $5$ and $x=5$ is the only solution. \[\dfrac{60}{5}+\dfrac{60}{10}+\dfrac{60}{15}+\dfrac{60}{20}=12+6+4+3=\boxed{25}.\]",251095320,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_24,1,"Mrs. Sanders has three grandchildren, who call her regularly. One calls her every three days, one calls her every four days, and one calls her every five days. All three called her on December 31, 2016. On how many days during the next year did she not receive a phone call from any of her grandchildren?1096$\textbf{(A) }78\qquad\textbf{(B) }80\qquad\textbf{(C) }144\qquad\textbf{(D) }146\qquad\textbf{(E) }152$","We use Principle of Inclusion-Exclusion. There are $365$ days in the year, and we subtract the days that she gets at least $1$ phone call, which is \[\left \lfloor \frac{365}{3} \right \rfloor +  \left \lfloor \frac{365}{4} \right \rfloor +  \left \lfloor \frac{365}{5} \right \rfloor.\]1097To this result we add the number of days where she gets at least $2$ phone calls in a day because we double subtracted these days, which is \[\left \lfloor \frac{365}{12} \right \rfloor +  \left \lfloor \frac{365}{15} \right \rfloor +  \left \lfloor \frac{365}{20} \right \rfloor.\]1098We now subtract the number of days where she gets three phone calls, which is $\left \lfloor \frac{365}{60} \right \rfloor.$ Therefore, our answer is \[365 - \left( \left \lfloor \frac{365}{3} \right \rfloor +  \left \lfloor \frac{365}{4} \right \rfloor +  \left \lfloor \frac{365}{5} \right \rfloor \right) +  \left( \left \lfloor \frac{365}{12} \right \rfloor +  \left \lfloor \frac{365}{15} \right \rfloor +  \left \lfloor \frac{365}{20} \right \rfloor \right) - \left \lfloor \frac{365}{60} \right \rfloor = 365 - 285+72 - 6 = \boxed{146}.\]",1461099321,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_24,2,"Mrs. Sanders has three grandchildren, who call her regularly. One calls her every three days, one calls her every four days, and one calls her every five days. All three called her on December 31, 2016. On how many days during the next year did she not receive a phone call from any of her grandchildren?1100$\textbf{(A) }78\qquad\textbf{(B) }80\qquad\textbf{(C) }144\qquad\textbf{(D) }146\qquad\textbf{(E) }152$","Note that $\operatorname{lcm}(3,4,5)=60,$ so there is a cycle every $60$ days.1101As shown below, all days in a cycle that Mrs. Sanders receives a phone call from any of her grandchildren are colored in red, yellow, or green. [asy] /* Made by MRENTHUSIASM */ size(7cm);  fill((2,6)--(3,6)--(3,5)--(2,5)--cycle,red); fill((5,6)--(6,6)--(6,5)--(5,5)--cycle,red); fill((8,6)--(9,6)--(9,5)--(8,5)--cycle,red); fill((1,5)--(2,5)--(2,4.5)--(1,4.5)--cycle,red); fill((4,5)--(5,5)--(5,4.5)--(4,4.5)--cycle,red); fill((7,5)--(8,5)--(8,4)--(7,4)--cycle,red); fill((0,4)--(1,4)--(1,3)--(0,3)--cycle,red); fill((3,4)--(4,4)--(4,3.5)--(3,3.5)--cycle,red); fill((6,4)--(7,4)--(7,3)--(6,3)--cycle,red); fill((9,4)--(10,4)--(10,3.5)--(9,3.5)--cycle,red); fill((2,3)--(3,3)--(3,2)--(2,2)--cycle,red); fill((5,3)--(6,3)--(6,2.5)--(5,2.5)--cycle,red); fill((8,3)--(9,3)--(9,2)--(8,2)--cycle,red); fill((1,2)--(2,2)--(2,1)--(1,1)--cycle,red); fill((4,2)--(5,2)--(5,1.5)--(4,1.5)--cycle,red); fill((7,2)--(8,2)--(8,1.5)--(7,1.5)--cycle,red); fill((0,1)--(1,1)--(1,0)--(0,0)--cycle,red); fill((3,1)--(4,1)--(4,0)--(3,0)--cycle,red); fill((6,1)--(7,1)--(7,0)--(6,0)--cycle,red); fill((9,1)--(10,1)--(10,2/3)--(9,2/3)--cycle,red); fill((3,6)--(4,6)--(4,5)--(3,5)--cycle,yellow); fill((7,6)--(8,6)--(8,5)--(7,5)--cycle,yellow); fill((1,4.5)--(2,4.5)--(2,4)--(1,4)--cycle,yellow); fill((5,5)--(6,5)--(6,4)--(5,4)--cycle,yellow); fill((9,5)--(10,5)--(10,4.5)--(9,4.5)--cycle,yellow); fill((3,3.5)--(4,3.5)--(4,3)--(3,3)--cycle,yellow); fill((7,4)--(8,4)--(8,3)--(7,3)--cycle,yellow); fill((1,3)--(2,3)--(2,2)--(1,2)--cycle,yellow); fill((5,2.5)--(6,2.5)--(6,2)--(5,2)--cycle,yellow); fill((9,3)--(10,3)--(10,2.5)--(9,2.5)--cycle,yellow); fill((3,2)--(4,2)--(4,1)--(3,1)--cycle,yellow); fill((7,1.5)--(8,1.5)--(8,1)--(7,1)--cycle,yellow); fill((1,1)--(2,1)--(2,0)--(1,0)--cycle,yellow); fill((5,1)--(6,1)--(6,0)--(5,0)--cycle,yellow); fill((9,2/3)--(10,2/3)--(10,1/3)--(9,1/3)--cycle,yellow); fill((4,6)--(5,6)--(5,5)--(4,5)--cycle,green); fill((9,6)--(10,6)--(10,5)--(9,5)--cycle,green); fill((4,4.5)--(5,4.5)--(5,4)--(4,4)--cycle,green); fill((9,4.5)--(10,4.5)--(10,4)--(9,4)--cycle,green); fill((4,4)--(5,4)--(5,3)--(4,3)--cycle,green); fill((9,3.5)--(10,3.5)--(10,3)--(9,3)--cycle,green); fill((4,3)--(5,3)--(5,2)--(4,2)--cycle,green); fill((9,2.5)--(10,2.5)--(10,2)--(9,2)--cycle,green); fill((4,1.5)--(5,1.5)--(5,1)--(4,1)--cycle,green); fill((9,2)--(10,2)--(10,1)--(9,1)--cycle,green); fill((4,1)--(5,1)--(5,0)--(4,0)--cycle,green); fill((9,1/3)--(10,1/3)--(10,0)--(9,0)--cycle,green);  real cur = 1; for (real i=6; i>0; --i) {    for (real j=0; j<10; ++j) {       label(""$""+string(cur)+""$"",(j+0.5,i-0.5));       ++cur;    } }  add(grid(10,6,linewidth(1.25))); [/asy] The year 2017 has $365$ days, or $6$ cycles and $5$ days.1102Together, the answer is $24\cdot6+2=\boxed{146}.$",1461103322,https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_24,3,"Mrs. Sanders has three grandchildren, who call her regularly. One calls her every three days, one calls her every four days, and one calls her every five days. All three called her on December 31, 2016. On how many days during the next year did she not receive a phone call from any of her grandchildren?1104$\textbf{(A) }78\qquad\textbf{(B) }80\qquad\textbf{(C) }144\qquad\textbf{(D) }146\qquad\textbf{(E) }152$","For any randomly chosen day, there is a $\frac{2}{3}$ chance the first child does not call her, a $\frac{3}{4}$ chance the second child does not call her and a $\frac{4}{5}$ chance the third child does not call her. So, in a randomly chosen day, there is a $\frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} = \frac{2}{5}$ chance no child calls her.1105In particular, this ""unrigorous"" reasoning becomes rigorous, by linearity of expectation, when we take a sample of $360$ days; over the first $360$ days, $360 \times \frac{2}{5} = 144$ days will go without a phone call. Now, we simply check the remaining days; on day $361$ and $362$ , nobody calls her; on day $363$ , the child that calls every three days will call her; on day $364$ , the child that calls every four days will call her; on day $365$ , the child that calls every five days will call her. Thus, we add two more days to $144$ , to get our answer of $\boxed{146}$",1461106323,https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_1,1,"The longest professional tennis match ever played lasted a total of $11$ hours and $5$ minutes. How many minutes was this?1107$\textbf{(A) }605\qquad\textbf{(B) }655\qquad\textbf{(C) }665\qquad\textbf{(D) }1005\qquad \textbf{(E) }1105$","It is best to split 11 hours and 5 minutes into 2 parts, one of 11 hours and another of 5 minutes.  We know that there is $60$ minutes in a hour.  Therefore, there are $11 \cdot 60 = 660$ minutes in 11 hours.  Adding the second part(the 5 minutes) we get $660 + 5 = \boxed{665}$",6651108324,https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_1,2,"The longest professional tennis match ever played lasted a total of $11$ hours and $5$ minutes. How many minutes was this?1109$\textbf{(A) }605\qquad\textbf{(B) }655\qquad\textbf{(C) }665\qquad\textbf{(D) }1005\qquad \textbf{(E) }1105$","The best method comes when you remember your multiplication tables. Thus trivial, we get our answer of $\boxed{665}$",6651110325,https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_1,3,"The longest professional tennis match ever played lasted a total of $11$ hours and $5$ minutes. How many minutes was this?1111$\textbf{(A) }605\qquad\textbf{(B) }655\qquad\textbf{(C) }665\qquad\textbf{(D) }1005\qquad \textbf{(E) }1105$","11 hours 5 min = $(11 \cdot 60) + 5 \text{min}  = 665 \text{min}$ , therefore $\boxed{665}$",6651112326,https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_2,1,"In rectangle $ABCD$ $AB=6$ and $AD=8$ . Point $M$ is the midpoint of $\overline{AD}$ . What is the area of $\triangle AMC$1113[asy]draw((0,4)--(0,0)--(6,0)--(6,8)--(0,8)--(0,4)--(6,8)--(0,0)); label(""$A$"", (0,0), SW); label(""$B$"", (6, 0), SE); label(""$C$"", (6,8), NE); label(""$D$"", (0, 8), NW); label(""$M$"", (0, 4), W); label(""$4$"", (0, 2), W); label(""$6$"", (3, 0), S);[/asy]1114$\textbf{(A) }12\qquad\textbf{(B) }15\qquad\textbf{(C) }18\qquad\textbf{(D) }20\qquad \textbf{(E) }24$","Using the triangle area formula for triangles: $A = \frac{bh}{2},$ where $A$ is the area, $b$ is the base, and $h$ is the height. This equation gives us $A = \frac{4 \cdot 6}{2} = \frac{24}{2} =\boxed{12}$",121115327,https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_2,2,"In rectangle $ABCD$ $AB=6$ and $AD=8$ . Point $M$ is the midpoint of $\overline{AD}$ . What is the area of $\triangle AMC$1116[asy]draw((0,4)--(0,0)--(6,0)--(6,8)--(0,8)--(0,4)--(6,8)--(0,0)); label(""$A$"", (0,0), SW); label(""$B$"", (6, 0), SE); label(""$C$"", (6,8), NE); label(""$D$"", (0, 8), NW); label(""$M$"", (0, 4), W); label(""$4$"", (0, 2), W); label(""$6$"", (3, 0), S);[/asy]1117$\textbf{(A) }12\qquad\textbf{(B) }15\qquad\textbf{(C) }18\qquad\textbf{(D) }20\qquad \textbf{(E) }24$","A triangle with the same height and base as a rectangle is half of the rectangle's area.  This means that a triangle with half of the base of the rectangle and also the same height means its area is one quarter of the rectangle's area.  Therefore, we get $\frac{48}{4} =\boxed{12}$",121118328,https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_2,3,"In rectangle $ABCD$ $AB=6$ and $AD=8$ . Point $M$ is the midpoint of $\overline{AD}$ . What is the area of $\triangle AMC$1119[asy]draw((0,4)--(0,0)--(6,0)--(6,8)--(0,8)--(0,4)--(6,8)--(0,0)); label(""$A$"", (0,0), SW); label(""$B$"", (6, 0), SE); label(""$C$"", (6,8), NE); label(""$D$"", (0, 8), NW); label(""$M$"", (0, 4), W); label(""$4$"", (0, 2), W); label(""$6$"", (3, 0), S);[/asy]1120$\textbf{(A) }12\qquad\textbf{(B) }15\qquad\textbf{(C) }18\qquad\textbf{(D) }20\qquad \textbf{(E) }24$","We can find the area of the entire rectangle, DCBA to be $8 \cdot 6=48$ and find DCM area to be $\frac{6 \cdot 4}{2} = 12$ and BCA to be $\frac{6 \cdot 8}{2}=24$ $48-12-24=$ $\boxed{12}$",121121329,https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_2,4,"In rectangle $ABCD$ $AB=6$ and $AD=8$ . Point $M$ is the midpoint of $\overline{AD}$ . What is the area of $\triangle AMC$1122[asy]draw((0,4)--(0,0)--(6,0)--(6,8)--(0,8)--(0,4)--(6,8)--(0,0)); label(""$A$"", (0,0), SW); label(""$B$"", (6, 0), SE); label(""$C$"", (6,8), NE); label(""$D$"", (0, 8), NW); label(""$M$"", (0, 4), W); label(""$4$"", (0, 2), W); label(""$6$"", (3, 0), S);[/asy]1123$\textbf{(A) }12\qquad\textbf{(B) }15\qquad\textbf{(C) }18\qquad\textbf{(D) }20\qquad \textbf{(E) }24$","A triangle is half of a rectangle. So since M is the midpoint of DA, we can see that triangle DAC is half of the whole rectangle. And it is also true that triangle MAC is half of triangle DAC, so triangle MAC would just be 1/4 of the whole rectangle. Since the rectangle's area is 48, 1/4 of 48 would be 12. Which gives us the answer $\boxed{12}$",121124330,https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_2,5,"In rectangle $ABCD$ $AB=6$ and $AD=8$ . Point $M$ is the midpoint of $\overline{AD}$ . What is the area of $\triangle AMC$1125[asy]draw((0,4)--(0,0)--(6,0)--(6,8)--(0,8)--(0,4)--(6,8)--(0,0)); label(""$A$"", (0,0), SW); label(""$B$"", (6, 0), SE); label(""$C$"", (6,8), NE); label(""$D$"", (0, 8), NW); label(""$M$"", (0, 4), W); label(""$4$"", (0, 2), W); label(""$6$"", (3, 0), S);[/asy]1126$\textbf{(A) }12\qquad\textbf{(B) }15\qquad\textbf{(C) }18\qquad\textbf{(D) }20\qquad \textbf{(E) }24$","We can subtract the total areas of triangles DCM and ABC from the rectangle ABCD. For triangle DCM, the base is 4 and the height is 6, so we multiply 4 and 6, then divide by 2 to get 12. For triangle ABC, the base is 4 and the height is 8, so we multiply 4 and 8, then divide by 2 to get 24. We add 24 and 12 to get 36. Then, we calculate the area of rectangle ABCD, which is 48. We subtract 36 from 48, resulting in $\boxed{12}$",121127331,https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_3,1,"Four students take an exam. Three of their scores are $70, 80,$ and $90$ . If the average of their four scores is $70$ , then what is the remaining score?1128$\textbf{(A) }40\qquad\textbf{(B) }50\qquad\textbf{(C) }55\qquad\textbf{(D) }60\qquad \textbf{(E) }70$","Let $r$ be the remaining student's score.  We know that the average, 70, is equal to $\frac{70 + 80 + 90 + r}{4}$ .  We can use basic algebra to solve for $r$ \[\frac{70 + 80 + 90 + r}{4} = 70\] \[\frac{240 + r}{4} = 70\] \[240 + r = 280\] \[r = 40\] giving us the answer of $\boxed{40}$",401129332,https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_3,2,"Four students take an exam. Three of their scores are $70, 80,$ and $90$ . If the average of their four scores is $70$ , then what is the remaining score?1130$\textbf{(A) }40\qquad\textbf{(B) }50\qquad\textbf{(C) }55\qquad\textbf{(D) }60\qquad \textbf{(E) }70$","Since $90$ is $20$ more than $70$ , and $80$ is $10$ more than $70$ , for $70$ to be the average, the other number must be $30$ less than $70$ , or $\boxed{40}$",401131333,https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_4,1,"When Cheenu was a boy, he could run $15$ miles in $3$ hours and $30$ minutes. As an old man, he can now walk $10$ miles in $4$ hours. How many minutes longer does it take for him to walk a mile now compared to when he was a boy?1132$\textbf{(A) }6\qquad\textbf{(B) }10\qquad\textbf{(C) }15\qquad\textbf{(D) }18\qquad \textbf{(E) }30$","When Cheenu was a boy, he could run $15$ miles in $3$ hours and $30$ minutes $= 3\times60 + 30$ minutes $= 210$ minutes, thus running $\frac{210}{15} = 14$ minutes per mile. Now that he is an old man, he can walk $10$ miles in $4$ hours $= 4 \times 60$ minutes $= 240$ minutes, thus walking $\frac{240}{10} = 24$ minutes per mile. Therefore, it takes him $\boxed{10}$ minutes longer to walk a mile now compared to when he was a boy.",101133334,https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_5,1,"The number $N$ is a two-digit number.1134• When $N$ is divided by $9$ , the remainder is $1$1135• When $N$ is divided by $10$ , the remainder is $3$1136What is the remainder when $N$ is divided by $11$1137$\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }7$","From the second bullet point, we know that the second digit must be $3$ , for a number divisible by $10$ ends in zero. Since there is a remainder of $1$ when $N$ is divided by $9$ , the multiple of $9$ must end in a $2$ for it to have the desired remainder $\pmod {10}.$ We now look for this one:1138$9(1)=9\\ 9(2)=18\\ 9(3)=27\\ 9(4)=36\\ 9(5)=45\\ 9(6)=54\\ 9(7)=63\\ 9(8)=72$1139The number $72+1=73$ satisfies both conditions. We subtract the biggest multiple of $11$ less than $73$ to get the remainder. Thus, $73-11(6)=73-66=\boxed{7}$",71140335,https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_5,2,"The number $N$ is a two-digit number.1141• When $N$ is divided by $9$ , the remainder is $1$1142• When $N$ is divided by $10$ , the remainder is $3$1143What is the remainder when $N$ is divided by $11$1144$\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }7$","This two digit number must take the form of $10x+y,$ where $x$ and $y$ are integers $0$ to $9.$ However, if x is an integer, we must have $y=3.$ So, the number's new form is $10x+3.$ This needs to have a remainder of $1$ when divided by $9.$ Because of the $9$ divisibility rule, we have \[10x+3 \equiv 1 \pmod 9.\] We subtract the three, getting \[10x \equiv -2 \pmod 9.\] which simplifies to \[10x \equiv 7 \pmod 9.\] However, $9x \equiv 0 \pmod 9,$ so \[10x - 9x \equiv 7 - 0 \pmod 9\] and \[x \equiv 7 \pmod 9.\]1145Let the quotient of $9$ in our modular equation be $c,$ and let our desired number be $z,$ so $x=9c+7$ and $z = 10x+3.$ We substitute these values into $z = 10x+3,$ and get \[z = 10(9c+7) + 3\] so \[z = 90c+73.\] As a result, $z \equiv 73 \pmod {90}.$1146To prove generalization vigorously, we can let $a$ be the remainder when $z$ is divided by $11.$ Setting up a modular equation, we have \[90c + 73 \equiv a \pmod {11}.\] Simplifying, \[90c+7 \equiv a \pmod {11}\] If $c = 1,$ then we don't have a 2 digit number! Thus, $c=0$ and $a=\boxed{7}$",71147336,https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_6,1,"The following bar graph represents the length (in letters) of the names of 19 people. What is the median length of these names? [asy] unitsize(0.9cm); draw((-0.5,0)--(10,0), linewidth(1.5)); draw((-0.5,1)--(10,1)); draw((-0.5,2)--(10,2)); draw((-0.5,3)--(10,3)); draw((-0.5,4)--(10,4)); draw((-0.5,5)--(10,5)); draw((-0.5,6)--(10,6)); draw((-0.5,7)--(10,7)); label(""frequency"",(-0.5,8)); label(""0"", (-1, 0)); label(""1"", (-1, 1)); label(""2"", (-1, 2)); label(""3"", (-1, 3)); label(""4"", (-1, 4)); label(""5"", (-1, 5)); label(""6"", (-1, 6)); label(""7"", (-1, 7)); filldraw((0,0)--(0,7)--(1,7)--(1,0)--cycle, black); filldraw((2,0)--(2,3)--(3,3)--(3,0)--cycle, black); filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, black); filldraw((6,0)--(6,4)--(7,4)--(7,0)--cycle, black); filldraw((8,0)--(8,4)--(9,4)--(9,0)--cycle, black); label(""3"", (0.5, -0.5)); label(""4"", (2.5, -0.5)); label(""5"", (4.5, -0.5)); label(""6"", (6.5, -0.5)); label(""7"", (8.5, -0.5)); label(""name length"", (4.5, -1)); [/asy]1148$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad \textbf{(E) }7$",We first notice that the median name will be the $(19+1)/2=10^{\mbox{th}}$ name. The $10^{\mbox{th}}$ name is $\boxed{4}$,41149337,https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_6,2,"The following bar graph represents the length (in letters) of the names of 19 people. What is the median length of these names? [asy] unitsize(0.9cm); draw((-0.5,0)--(10,0), linewidth(1.5)); draw((-0.5,1)--(10,1)); draw((-0.5,2)--(10,2)); draw((-0.5,3)--(10,3)); draw((-0.5,4)--(10,4)); draw((-0.5,5)--(10,5)); draw((-0.5,6)--(10,6)); draw((-0.5,7)--(10,7)); label(""frequency"",(-0.5,8)); label(""0"", (-1, 0)); label(""1"", (-1, 1)); label(""2"", (-1, 2)); label(""3"", (-1, 3)); label(""4"", (-1, 4)); label(""5"", (-1, 5)); label(""6"", (-1, 6)); label(""7"", (-1, 7)); filldraw((0,0)--(0,7)--(1,7)--(1,0)--cycle, black); filldraw((2,0)--(2,3)--(3,3)--(3,0)--cycle, black); filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, black); filldraw((6,0)--(6,4)--(7,4)--(7,0)--cycle, black); filldraw((8,0)--(8,4)--(9,4)--(9,0)--cycle, black); label(""3"", (0.5, -0.5)); label(""4"", (2.5, -0.5)); label(""5"", (4.5, -0.5)); label(""6"", (6.5, -0.5)); label(""7"", (8.5, -0.5)); label(""name length"", (4.5, -1)); [/asy]1150$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad \textbf{(E) }7$","To find the median length of a name from a bar graph, we must add up the number of names. Doing so gives us $7 + 3 + 1 + 4 + 4 = 19$ . Thus the index of the median length would be the 10th name. Since there are $7$ names with length $3$ , and $3$ names with length $4$ , the $10$ th name would have $4$ letters. Thus our answer is $\boxed{4}$",41151338,https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_8,1,Find the value of the expression \[100-98+96-94+92-90+\cdots+8-6+4-2.\] $\textbf{(A) }20\qquad\textbf{(B) }40\qquad\textbf{(C) }50\qquad\textbf{(D) }80\qquad \textbf{(E) }100$,"We can group each subtracting pair together: \[(100-98)+(96-94)+(92-90)+ \ldots +(8-6)+(4-2).\] After subtracting, we have: \[2+2+2+\ldots+2+2=2(1+1+1+\ldots+1+1).\] There are $50$ even numbers, therefore there are $\dfrac{50}{2}=25$ even pairs. Therefore the sum is $2 \cdot 25=\boxed{50}$",501152339,https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_8,2,Find the value of the expression \[100-98+96-94+92-90+\cdots+8-6+4-2.\] $\textbf{(A) }20\qquad\textbf{(B) }40\qquad\textbf{(C) }50\qquad\textbf{(D) }80\qquad \textbf{(E) }100$,"Since our list does not end with one, we divide every number by 2 and we end up with \[50-49+48-47+ \ldots +4-3+2-1\] We can group each subtracting pair together: \[(50-49)+(48-47)+(46-45)+ \ldots +(4-3)+(2-1).\] There are now $25$ pairs of numbers, and the value of each pair is $1$ .  This sum is $25$ . However, we divided by $2$ originally so we will multiply $2*25$ to get the final answer of $\boxed{50}$",501153340,https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_9,1,"What is the sum of the distinct prime integer divisors of $2016$1154$\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }16\qquad\textbf{(D) }49\qquad \textbf{(E) }63$","The prime factorization is $2016=2^5\times3^2\times7$ .  Since the problem is only asking us for the distinct prime factors, we have $2,3,7$ .  Their desired sum is then $\boxed{12}$",121155341,https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_9,2,"What is the sum of the distinct prime integer divisors of $2016$1156$\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }16\qquad\textbf{(D) }49\qquad \textbf{(E) }63$","We notice that $9 \mid 2016$ , since $2+0+1+6 = 9$ , and $9 \mid 9$ . We can divide $2016$ by $9$ to get $224$ . This is divisible by $4$ , as $4 \mid 24$ . Dividing $224$ by $4$ , we have $56$ . This is clearly divisible by $7$ , leaving $8$ . We have $2016 = 9\cdot 4\cdot 7\cdot 8$ . We know that $4$ and $8$ are both multiples of $2$ $9$ is $3^2$ , and $7$ is prime. This means that the distinct prime factors are $2,3,$ and $7$ . Their sum is $\boxed{12}$",121157342,https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_10,1,Suppose that $a * b$ means $3a-b.$ What is the value of $x$ if \[2 * (5 * x)=1\] $\textbf{(A) }\frac{1}{10} \qquad\textbf{(B) }2\qquad\textbf{(C) }\frac{10}{3} \qquad\textbf{(D) }10\qquad \textbf{(E) }14$,"Let us plug in $(5 * x)=1$ into $3a-b$ . Thus it would be $3(5)-x$ . Now we have $2*(15-x)=1$ . Plugging $2*(15-x)$ into $3a-b$ , we have $6-15+x=1$ . Solving for $x$ we have \[-9+x=1\] \[x=\boxed{10}\]",101158343,https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_10,2,Suppose that $a * b$ means $3a-b.$ What is the value of $x$ if \[2 * (5 * x)=1\] $\textbf{(A) }\frac{1}{10} \qquad\textbf{(B) }2\qquad\textbf{(C) }\frac{10}{3} \qquad\textbf{(D) }10\qquad \textbf{(E) }14$,"Let us set a variable $y$ equal to $5 * x$ . Solving for y in the equation $3(2)-y=1$ , we see that y is equal to five. By substitution, we see that $5 * x$ = 5. Solving for x in the equation $5(3)-x = 5$ we get \[x=\boxed{10}\]",101159344,https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_11,1,"Determine how many two-digit numbers satisfy the following property: when the number is added to the number obtained by reversing its digits, the sum is $132.$1160$\textbf{(A) }5\qquad\textbf{(B) }7\qquad\textbf{(C) }9\qquad\textbf{(D) }11\qquad \textbf{(E) }12$","We can write the two digit number in the form of $10a+b$ ; reverse of $10a+b$ is $10b+a$ . The sum of those numbers is: \[(10a+b)+(10b+a)=132\] \[11a+11b=132\] \[a+b=12\] We can use brute force to find order pairs $(a,b)$ such that $a+b=12$ . Since $a$ and $b$ are both digits, both $a$ and $b$ have to be integers less than $10$ . Thus, our ordered pairs are $(3,9); (4,8); (5,7); (6,6); (7,5); (8,4); (9,3)$ ; or $\boxed{7}$ ordered pairs.",71161345,https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_11,2,"Determine how many two-digit numbers satisfy the following property: when the number is added to the number obtained by reversing its digits, the sum is $132.$1162$\textbf{(A) }5\qquad\textbf{(B) }7\qquad\textbf{(C) }9\qquad\textbf{(D) }11\qquad \textbf{(E) }12$","Since the numbers are “mirror images,” their average has to be $\frac{132}{2}=66$ . The highest possible value for the tens digit is $9$ because it is a two-digit number. $9-6=3$ and $6-3=3$ , so our lowest tens digit is $3$ . The numbers between $9$ and $3$ inclusive is $9-3+1=\boxed{7}$ total possibilities.",71163346,https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_14,1,"Karl's car uses a gallon of gas every $35$ miles, and his gas tank holds $14$ gallons when it is full. One day, Karl started with a full tank of gas, drove $350$ miles, bought $8$ gallons of gas, and continued driving to his destination. When he arrived, his gas tank was half full. How many miles did Karl drive that day?1164$\textbf{(A) }525\qquad\textbf{(B) }560\qquad\textbf{(C) }595\qquad\textbf{(D) }665\qquad \textbf{(E) }735$","Since he uses a gallon of gas every $35$ miles, he had used $\frac{350}{35} = 10$ gallons after $350$ miles. Therefore, after the first leg of his trip he had $14 - 10 = 4$ gallons of gas left. Then, he bought $8$ gallons of gas, which brought him up to $12$ gallons of gas in his gas tank. When he arrived, he had $\frac{1}{2} \cdot 14 = 7$ gallons of gas. So he used $5$ gallons of gas on the second leg of his trip. Therefore,                                               the second part of his trip covered $5 \cdot 35 = 175$ miles. Adding this to the $350$ miles, we see that he drove $350 + 175 = \boxed{525}$ miles.",5251165347,https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_15,1,"What is the largest power of $2$ that is a divisor of $13^4 - 11^4$1166$\textbf{(A)}\mbox{ }8\qquad \textbf{(B)}\mbox{ }16\qquad \textbf{(C)}\mbox{ }32\qquad \textbf{(D)}\mbox{ }64\qquad \textbf{(E)}\mbox{ }128$","First, we use difference of squares on $13^4 - 11^4 = (13^2)^2 - (11^2)^2$ to get $13^4 - 11^4 = (13^2 + 11^2)(13^2 - 11^2)$ . Using difference of squares again and simplifying, we get $(169 + 121)(13+11)(13-11) = 290 \cdot 24 \cdot 2 = (2\cdot 8 \cdot 2) \cdot (3 \cdot 145)$ . Realizing that we don't need the right-hand side because it doesn't contain any factor of 2, we see that the greatest power of $2$ that is a divisor $13^4 - 11^4$ is $\boxed{32}$",321167348,https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_15,2,"What is the largest power of $2$ that is a divisor of $13^4 - 11^4$1168$\textbf{(A)}\mbox{ }8\qquad \textbf{(B)}\mbox{ }16\qquad \textbf{(C)}\mbox{ }32\qquad \textbf{(D)}\mbox{ }64\qquad \textbf{(E)}\mbox{ }128$","Just like in the above solution, we use the difference-of-squares factorization, but only once to get $13^4-11^4=(13^2-11^2)(13^2+11^2).$ We can then compute that this is equal to $48\cdot290.$ Note that $290=2\cdot145$ (we don't need to factorize any further as $145$ is already odd) thus the largest power of $2$ that divides $290$ is only $2^1=2,$ while $48=2^4\cdot3,$ so the largest power of $2$ that divides $48$ is $2^4=16.$ Hence, the largest power of $2$ that is a divisor of $13^4-11^4$ is $2\cdot16=\boxed{32}.$",321169349,https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_15,3,"What is the largest power of $2$ that is a divisor of $13^4 - 11^4$1170$\textbf{(A)}\mbox{ }8\qquad \textbf{(B)}\mbox{ }16\qquad \textbf{(C)}\mbox{ }32\qquad \textbf{(D)}\mbox{ }64\qquad \textbf{(E)}\mbox{ }128$","Let $n=13^4-11^4.$ We wish to find the largest power of $2$ that divides $n$1171Denote $v_p(k)$ as the largest exponent of $p$ in the prime factorization of $n$ . In this problem, we have $p=2$1172By the Lifting the Exponent Lemma on $n$1173\[v_2(13^4-11^4)=v_2(13-11)+v_2(4)+v_2(13+11)-1\] \[=v_2(2)+v_2(4)+v_2(24)-1\] \[=1+2+3-1=5.\]1174Therefore, exponent of the largest power of $2$ that divids $13^4-11^4$ is $5,$ so the largest power of $2$ that divides this number is $2^5=\boxed{32}$",321175350,https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_15,4,"What is the largest power of $2$ that is a divisor of $13^4 - 11^4$1176$\textbf{(A)}\mbox{ }8\qquad \textbf{(B)}\mbox{ }16\qquad \textbf{(C)}\mbox{ }32\qquad \textbf{(D)}\mbox{ }64\qquad \textbf{(E)}\mbox{ }128$","We can simply take 13 to the 4th power, which is 28561. We subtract that by 11 to the 4th power, which is 14641 (You can use Pascal's Triangle to find this). Finally, subtract the numbers to get 13920.1177To test the options, since we need the largest one, we can go from top down. Testing, we see that both D and E are decimals,1178and 32 works. So, our answer is $\boxed{32}.$",321179351,https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_16,1,"Annie and Bonnie are running laps around a $400$ -meter oval track. They started together, but Annie has pulled ahead, because she runs $25\%$ faster than Bonnie. How many laps will Annie have run when she first passes Bonnie?1180$\textbf{(A) }1\dfrac{1}{4}\qquad\textbf{(B) }3\dfrac{1}{3}\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }25$","Each lap Bonnie runs, Annie runs another quarter lap, so Bonnie will run four laps before she is overtaken.  This means that Annie and Bonnie are equal so that Annie needs to run another lap to overtake Bonnie. That means Annie will have run $\boxed{5}$ laps.",51181352,https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_16,2,"Annie and Bonnie are running laps around a $400$ -meter oval track. They started together, but Annie has pulled ahead, because she runs $25\%$ faster than Bonnie. How many laps will Annie have run when she first passes Bonnie?1182$\textbf{(A) }1\dfrac{1}{4}\qquad\textbf{(B) }3\dfrac{1}{3}\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }25$","Call $x$ the distance Annie runs. If Annie is $25\%$ faster than Bonnie, then Bonnie will run a distance of $\frac{4}{5}x$ . For Annie to meet Bonnie, she must run an extra $400$ meters, the length of the track. So $x-\left(\frac{4}{5}\right)x=400 \implies x=2000$ , which is $\boxed{5}$ laps.",51183353,https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_17,1,"An ATM password at Fred's Bank is composed of four digits from $0$ to $9$ , with repeated digits allowable. If no password may begin with the sequence $9,1,1,$ then how many passwords are possible?1184$\textbf{(A)}\mbox{ }30\qquad\textbf{(B)}\mbox{ }7290\qquad\textbf{(C)}\mbox{ }9000\qquad\textbf{(D)}\mbox{ }9990\qquad\textbf{(E)}\mbox{ }9999$","For the first three digits, there are $10^3-1=999$ combinations since $911$ is not allowed. For the final digit, any of the $10$ numbers are allowed. $999 \cdot 10 = 9990 \rightarrow \boxed{9990}$",9901185354,https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_17,2,"An ATM password at Fred's Bank is composed of four digits from $0$ to $9$ , with repeated digits allowable. If no password may begin with the sequence $9,1,1,$ then how many passwords are possible?1186$\textbf{(A)}\mbox{ }30\qquad\textbf{(B)}\mbox{ }7290\qquad\textbf{(C)}\mbox{ }9000\qquad\textbf{(D)}\mbox{ }9990\qquad\textbf{(E)}\mbox{ }9999$","Counting the prohibited cases, we find that there are 10 of them. This is because, when we start with 9,1, and 1, we can have any of the 10 digits for the last digit. So, our answer is $10^4-10=\boxed{9990}.$",9901187355,https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_18,1,"In an All-Area track meet, $216$ sprinters enter a $100-$ meter dash competition. The track has $6$ lanes, so only $6$ sprinters can compete at a time. At the end of each race, the five non-winners are eliminated, and the winner will compete again in a later race. How many races are needed to determine the champion sprinter?1188$\textbf{(A)}\mbox{ }36\qquad\textbf{(B)}\mbox{ }42\qquad\textbf{(C)}\mbox{ }43\qquad\textbf{(D)}\mbox{ }60\qquad\textbf{(E)}\mbox{ }72$","From any $n-$ th race, only $\frac{1}{6}$ will continue on. Since we wish to find the total number of races, a column representing the races over time is ideal.1189Starting with the first race: \[\frac{216}{6}=36\] \[\frac{36}{6}=6\] \[\frac{6}{6}=1\] Adding all of the numbers in the second column yields $\boxed{43}$",431190356,https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_18,2,"In an All-Area track meet, $216$ sprinters enter a $100-$ meter dash competition. The track has $6$ lanes, so only $6$ sprinters can compete at a time. At the end of each race, the five non-winners are eliminated, and the winner will compete again in a later race. How many races are needed to determine the champion sprinter?1191$\textbf{(A)}\mbox{ }36\qquad\textbf{(B)}\mbox{ }42\qquad\textbf{(C)}\mbox{ }43\qquad\textbf{(D)}\mbox{ }60\qquad\textbf{(E)}\mbox{ }72$","Every race eliminates $5$ players. The winner is decided when there is only $1$ runner left. You can construct the equation: $216$ $5x$ $1$ . Thus, $215$ players have to be eliminated. Therefore, we need $\frac{215}{5}$ games to decide the winner, or $\boxed{43}$",431192357,https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_19,1,"The sum of $25$ consecutive even integers is $10,000$ . What is the largest of these $25$ consecutive integers?1193$\textbf{(A)}\mbox{ }360\qquad\textbf{(B)}\mbox{ }388\qquad\textbf{(C)}\mbox{ }412\qquad\textbf{(D)}\mbox{ }416\qquad\textbf{(E)}\mbox{ }424$","Let $n$ be the 13th consecutive even integer that's being added up. By doing this, we can see that the sum of all 25 even numbers will simplify to $25n$ since $(n-2k)+\dots+(n-4)+(n-2)+(n)+(n+2)+(n+4)+ \dots +(n+2k)=25n$ . Now, $25n=10000 \rightarrow n=400$ . Remembering that this is the 13th integer, we wish to find the 25th, which is $400+2(25-13)=\boxed{424}$",4241194358,https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_19,2,"The sum of $25$ consecutive even integers is $10,000$ . What is the largest of these $25$ consecutive integers?1195$\textbf{(A)}\mbox{ }360\qquad\textbf{(B)}\mbox{ }388\qquad\textbf{(C)}\mbox{ }412\qquad\textbf{(D)}\mbox{ }416\qquad\textbf{(E)}\mbox{ }424$","Let $x$ be the largest number. Then, $x+(x-2)+(x-4)+\cdots +(x-48)=10000$ . Factoring this gives $2\left(\frac{x}{2} + \left(\frac{x}{2} - 1\right) + \left(\frac{x}{2} - 2\right) +\cdots + \left(\frac{x}{2} - 24\right)\right)=10,000$ . Grouping like terms gives $25\left(\frac{x}{2}\right) - 300=5000$ , and continuing down the line, we find $x=\boxed{424}$",4241196359,https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_19,3,"The sum of $25$ consecutive even integers is $10,000$ . What is the largest of these $25$ consecutive integers?1197$\textbf{(A)}\mbox{ }360\qquad\textbf{(B)}\mbox{ }388\qquad\textbf{(C)}\mbox{ }412\qquad\textbf{(D)}\mbox{ }416\qquad\textbf{(E)}\mbox{ }424$","Let $x$ be the smallest number. The equation will become, $x+(x+2)+(x+4)+\cdots +(x+48)=10,000$ . After you combine like terms, you get $25x+(50*12)=10,000$ which turns into $10,000-600=25x$ $25x=9400$ , so $x=376$ . Then, you add $376+48 = \boxed{424}$",4241198360,https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_19,4,"The sum of $25$ consecutive even integers is $10,000$ . What is the largest of these $25$ consecutive integers?1199$\textbf{(A)}\mbox{ }360\qquad\textbf{(B)}\mbox{ }388\qquad\textbf{(C)}\mbox{ }412\qquad\textbf{(D)}\mbox{ }416\qquad\textbf{(E)}\mbox{ }424$","Dividing the series by $2$ , we get that the sum of $25$ consecutive integers is $5000$ . Let the middle number be $k$ we know that the sum is $25k$ , so $25k=5000$ . Solving, $k=200$ $2k=400$ is the middle term of the original sequence, so the original last term is $400+\frac{25-1}{2}\cdot 2=424$ . So the answer is $\boxed{424}$",4241200361,https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_20,1,"The least common multiple of $a$ and $b$ is $12$ , and the least common multiple of $b$ and $c$ is $15$ . What is the least possible value of the least common multiple of $a$ and $c$

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