Question

设 $N$ 是最小的正整数，它既比某个整数少 $22$ %，又比另一个整数多 $16$ %。求 $N$ 除以 $1000$ 的余数。

Let $N$ be the least positive integer that is both $22$ percent less than one integer and $16$ percent greater than another integer. Find the remainder when $N$ is divided by $1000.$

答案

答案：131

知识点：百分数整除性最小公倍数

难度评级：2050

小提示：

写成 $N = \frac{78}{100}a = \frac{116}{100}b$ ，并把两个分数都约到最简

Write $N = \frac{78}{100}a = \frac{116}{100}b$ and reduce both fractions to lowest terms

大提示：

最简形式为 $N = \frac{39}{50}a = \frac{29}{25}b$ ，这会迫使 $N$ 同时是 $39$ 和 $29$ 的倍数

In lowest terms $N = \frac{39}{50}a = \frac{29}{25}b,$ which forces $N$ to be a multiple of both $39$ and $29$

解答：

条件说明对某些整数 $a$ 和 $b$ 有 $N = \frac{78}{100}a = \frac{39}{50}a$ 且 $N = \frac{116}{100}b = \frac{29}{25}b$ 。因为 $\gcd(39, 50) = 1$ ，第一个等式迫使 $a$ 能被 $50$ 整除，所以 $N$ 是 $39$ 的倍数；因为 $\gcd(29, 25) = 1$ ，第二个等式迫使 $b$ 能被 $25$ 整除，所以 $N$ 是 $29$ 的倍数。

同时被这两个数整除的最小正整数是 $N = 39 \cdot 29 = 1131$ ，取 $a = 1450$ 和 $b = 975$ 时可以达到。除以 $1000$ 的余数为 $131$ 。

The conditions say $N = \frac{78}{100}a = \frac{39}{50}a$ and $N = \frac{116}{100}b = \frac{29}{25}b$ for some integers $a$ and $b.$ Since $\gcd(39, 50) = 1,$ the first equation forces $a$ to be divisible by $50,$ so $N$ is a multiple of $39;$ since $\gcd(29, 25) = 1,$ the second forces $b$ to be divisible by $25,$ so $N$ is a multiple of $29.$

The least positive integer divisible by both is $N = 39 \cdot 29 = 1131,$ achieved with $a = 1450$ and $b = 975.$ The remainder upon division by $1000$ is $131.$

2015 AIME II 第 1 题

答案：131

难度评级：2050

小提示：

大提示：

解答：

其他年份的第 1 题