Question

某高中的 $2001$ 名学生每人都学习西班牙语或法语，也有一些人两种语言都学。学习西班牙语的人数占全校人数的百分之 $80$ 到百分之 $85$ 之间，学习法语的人数占百分之 $30$ 到百分之 $40$ 之间。设 $m$ 为可能同时学习两种语言的最少学生数， $M$ 为可能同时学习两种语言的最多学生数。求 $M - m$ 。

Each of the $2001$ students at a high school studies either Spanish or French, and some study both. The number who study Spanish is between $80$ percent and $85$ percent of the school population, and the number who study French is between $30$ percent and $40$ percent. Let $m$ be the smallest number of students who could study both languages, and let $M$ be the largest number of students who could study both languages. Find $M - m.$

答案

答案：298

知识点：容斥原理百分数极限情形界定

难度评级：2110

解答：

设学习西班牙语和法语的人数分别为 $s$ 和 $f$ 。因为每个学生至少学习一种语言，同时学习两种语言的人数为 $s + f - 2001$ 。由范围 $1600.8 \lt s \lt 1700.85$ 得 $1601 \le s \le 1700$ ，由 $600.3 \lt f \lt 800.4$ 得 $601 \le f \le 800$ 。

重叠人数在 $s + f$ 最小时最小，所以 $m = 1601 + 601 - 2001 = 201$ ；在 $s + f$ 最大时最大，所以 $M = 1700 + 800 - 2001 = 499$ 。这两个极端都可以达到，因此 $M - m = 499 - 201 = 298$ 。

Let $s$ and $f$ be the numbers of students studying Spanish and French. Since every student studies at least one language, the number studying both is $s + f - 2001.$ The bounds $1600.8 \lt s \lt 1700.85$ force $1601 \le s \le 1700,$ and $600.3 \lt f \lt 800.4$ force $601 \le f \le 800.$

The overlap is smallest when $s + f$ is smallest, giving $m = 1601 + 601 - 2001 = 201,$ and largest when $s + f$ is largest, giving $M = 1700 + 800 - 2001 = 499.$ Both extremes are achievable, so $M - m = 499 - 201 = 298.$

2001 AIME II 第 2 题

答案：298

难度评级：2110

解答：

其他年份的第 2 题