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math-ai/BlueMO

BlueMO 🚀 BlueMO: A Comprehensive Collection of Challenging Mathematical Olympiad Problems from the Little Blue Book Series   BlueMO is a comprehensive and challenging dataset comprising mathematical olympiad problems paired with detailed solutions, meticulously curated from the esteemed "Little Blue Book" (小蓝书) series (Second Edition)—a vital resource for Chinese students training for national and international olympiad math competitions. Designed to… See the full description on the dataset page: https://huggingface.co/datasets/math-ai/BlueMO.

sourceHugging Facecc-by-nd-4.0updated 8mo agoView on Hugging Face
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1{2    "source_file": "./raw_volume-zh/volume5/exercise2.tex",3    "problem_type": "calculation",4    "problem": "问题4. 求所有的正整数 $a_1, a_2, \\cdots, a_n$, 使得\n$$\n\\frac{99}{100}=\\frac{a_0}{a_1}+\\frac{a_1}{a_2}+\\cdots+\\frac{a_{n-1}}{a_n}\n$$\n其中 $a_0=1$, 且 $\\left(a_{k+1}-1\\right) a_{k-1} \\geqslant a_k^2-1(k=1,2, \\cdots, n-1)$.",5    "solution": "显然 $a_k>a_{k-1}$, 且 $a_k \\geqslant 2 . k=1,2, \\cdots, n-1$.\n由条件可得\n$$\n\\frac{a_{k-1}}{a_k} \\leqslant \\frac{a_{k-1}}{a_k-1}-\\frac{a_k}{a_{k+1}-1} .\n$$\n上面不等式对 $k=i+1, i+2, \\cdots, n$ 求和, 得\n$$\n\\frac{a_i}{a_{i+1}}+\\frac{a_{i+1}}{a_{i+2}}+\\cdots+\\frac{a_{n-1}}{a_n}<\\frac{a_i}{a_{i+1}-1} .\n$$\n当 $i=0$ 时, 由题设及上式, $\\frac{1}{a_1} \\leqslant \\frac{99}{100}<\\frac{1}{a_1-1}$. 则\n$$\n\\frac{100}{99} \\leqslant a_1<\\frac{100}{99}+1 \\text {, 故 } a_1=2 \\text {. }\n$$\n当 $i=1 、 2 、 3$ 时, 同样可求得 $a_2=5, a_3=56, a_4=78400$.\n所以 $\\frac{1}{a_5} \\leqslant \\frac{1}{a_4}\\left(\\frac{99}{100}-\\frac{1}{2}-\\frac{2}{5}-\\frac{5}{56}-\\frac{56}{25-56^2}\\right)=0$, 故没有 $a_5$, 因此, 所求的正整数 $a_1=2, a_2=5, a_3=56, a_4=78400$.",6    "remark": "",7    "figures": []8}