Question

令 $S$ 为所有形如 $\frac{a}{b}$ 的数之和，其中 $a$ 和 $b$ 是 $1000$ 的互质正因数。求不超过 $\frac{S}{10}$ 的最大整数。

Let $S$ be the sum of all numbers of the form $\frac{a}{b},$ where $a$ and $b$ are relatively prime positive divisors of $1000.$ What is the greatest integer that does not exceed $\frac{S}{10}?$

答案

答案：248

知识点：因数质因数分解等比数列

难度评级：2450

解答：

写 $a = 2^i 5^j$ 、 $b = 2^k 5^l$ ，其中各指数均在 $0$ 到 $3$ 之间。互质意味着 $\min(i, k) = 0$ 且 $\min(j, l) = 0$ ，这两个条件彼此独立。当 $(a, b)$ 遍历所有互质数对时，因子 $2^{i - k}$ 恰好遍历 $\{2^{-3}, \ldots, 2^{3}\}$ 中的每个值， $5^{j - l}$ 也同理。因此 $\begin{aligned} S &= \scriptsize \left(1 + 2 + 4 + 8 + \tfrac{1}{2} + \tfrac{1}{4} + \tfrac{1}{8}\right) \\ &\quad \scriptsize {}\cdot \left(1 + 5 + 25 + 125 + \tfrac{1}{5} + \tfrac{1}{25} + \tfrac{1}{125}\right) \\ &= \frac{127}{8} \cdot \frac{19531}{125}. \end{aligned}$

这等于 $\frac{2480437}{1000} = 2480.437$ ，所以 $\frac{S}{10} = 248.0437$ 。不超过它的最大整数是 $248$ 。

Write $a = 2^i 5^j$ and $b = 2^k 5^l$ with exponents between $0$ and $3.$ Coprimality means $\min(i, k) = 0$ and $\min(j, l) = 0,$ and these two constraints are independent. So as $(a, b)$ runs over all coprime pairs, the factor $2^{i - k}$ independently takes each value in $\{2^{-3}, \ldots, 2^{3}\}$ exactly once, and similarly for $5^{j - l}.$ Hence $\begin{aligned} S &= \scriptsize \left(1 + 2 + 4 + 8 + \tfrac{1}{2} + \tfrac{1}{4} + \tfrac{1}{8}\right) \\ &\quad \scriptsize {}\cdot \left(1 + 5 + 25 + 125 + \tfrac{1}{5} + \tfrac{1}{25} + \tfrac{1}{125}\right) \\ &= \frac{127}{8} \cdot \frac{19531}{125}. \end{aligned}$

This equals $\frac{2480437}{1000} = 2480.437,$ so $\frac{S}{10} = 248.0437,$ and the greatest integer not exceeding it is $248.$

2000 AIME I 第 11 题

答案：248

难度评级：2450

解答：

其他年份的第 11 题