Question

求有多少个有理数 $r$ ，满足 $0 \lt r \lt 1$ ，并且当 $r$ 写成最简分数时，分子和分母之和为 $1000$ 。

Find the number of rational numbers $r,$ $0 \lt r \lt 1,$ such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of $1000.$

答案

答案：200

知识点：欧拉函数最大公约数分数

难度评级：2110

解答：

将 $r = \frac{a}{b}$ 写成最简分数，且 $a + b = 1000$ ；由于 $0 \lt r \lt 1$ ，需要 $1 \le a \le 499$ 。又因为 $\gcd(a, b) = \gcd(a, 1000 - a)$ $= \gcd(a, 1000)$ ，这个分数为最简分数当且仅当 $a$ 与 $1000$ 互质。

在 $[1, 999]$ 中与 $1000$ 互质的整数有 $\varphi(1000) = 1000 \cdot \frac{1}{2} \cdot \frac{4}{5} = 400$ 个，它们按 $a \leftrightarrow 1000 - a$ 成对出现（注意 $a = 500$ 与 $1000$ 不互质），所以其中恰有 $200$ 个小于 $500$ 。答案为 $200$ 。

Write $r = \frac{a}{b}$ in lowest terms with $a + b = 1000;$ since $0 \lt r \lt 1,$ we need $1 \le a \le 499.$ Because $\gcd(a, b) = \gcd(a, 1000 - a)$ $= \gcd(a, 1000),$ the fraction is in lowest terms exactly when $a$ is coprime to $1000.$

There are $\varphi(1000) = 1000 \cdot \frac{1}{2} \cdot \frac{4}{5} = 400$ integers in $[1, 999]$ coprime to $1000,$ and they pair up as $a \leftrightarrow 1000 - a$ (note $a = 500$ is not coprime to $1000$ ), so exactly $200$ of them are less than $500.$ The answer is $200.$

2014 AIME I 第 3 题

答案：200

难度评级：2110

解答：

其他年份的第 3 题