CoolFace
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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/volume14/exercise7.tex",3    "problem_type": "calculation",4    "problem": "问题5. 设正整数 $n$ 的所有正约数从小到大依次为 $d_1<d_2<\\cdots<d_k(k \\geqslant 4)$, 且满足 $d_1^2+d_2^2+d_3^2+d_4^2=n$, 求 $n$ 的值.",5    "solution": "若 $n$ 为奇数,则所有正约数均为奇数, 从而 $d_1^2+d_2^2+d_3^2+d_4^2$ 为偶数, 不可能等于 $n$. 故 $n$ 为偶数, 从而 $d_1=1, d_2=2$.\n如果 $n$ 是 4 的倍数, 那么 $d_3, d_4$ 中有一个等于 4 , 另一个必须为奇数, 此时 $d_1^2+d_2^2+d_3^2+d_4^2 \\equiv 2(\\bmod 4)$, 不可能等于 $n$. 因此 $n=2 m, m$ 为奇数.\n显然 $d_3$ 为 $m$ 的最小奇质因数,则 $d_4$ 为偶数,故 $d_4$ 必为除 2 之外 $n$ 的最小偶约数, 从而 $d_4=2 d_3$.\n以下可知 $n=1^2+2^2+d_3^2+4 d_3^2=5\\left(1+d_3^2\\right)$, 故 5 是 $n$ 的约数, 因此 $d_3$, $d_4$ 不可能是 3 和 6 , 只能是 5 和 10 , 于是 $n=1^2+2^2+d_3^2+4 d_3^2=5(1+ \\left.5^2\\right)=130$.\n经检验, $n=130$ 满足题意.",6    "remark": "",7    "figures": []8}