1987 AIME 第 3 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

3.

一个自然数的真因数是除 11 和该数本身以外的正整数因数。如果一个大于 11 的自然数等于其所有不同真因数的乘积,就称它为“美好数”。求前十个美好数之和。

A proper divisor of a natural number is a positive integral divisor other than 11 and the number itself. A natural number greater than 11 is called “nice” if it equals the product of its distinct proper divisors. What is the sum of the first ten nice numbers?

答案:182
知识点:因数个数质因数分解
难度评级:1830
小提示:

用该数及其正因数个数表示所有正因数的乘积

Express the product of all positive divisors in terms of the number and its divisor count

大提示:

美好数恰好是有四个正因数的数

The nice numbers are exactly those having four positive divisors

解答:

NNdd 个正因数,则它们的乘积为 Nd2N^{\frac{d}{2}}。去掉 11NN 后,剩余乘积为 Nd21N^{\frac{d}{2}-1},它恰好在 d=4d=4 时等于 NN。因此,美好数恰好是 p3p^3pqpq 形式的数,其中后者的两个质数不同。前十个美好数为 668810101414151521212222262627273333,它们的和为 182182

If NN has dd positive divisors, their product is Nd2.N^{\frac{d}{2}}. Removing 11 and NN leaves product Nd21,N^{\frac{d}{2}-1}, which equals NN exactly when d=4.d=4. Thus nice numbers are precisely p3p^3 and pqpq for distinct primes. The first ten are 6,6, 8,8, 10,10, 14,14, 15,15, 21,21, 22,22, 26,26, 27,27, 33,33, whose sum is 182.182.

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