AIME 2001 II · 第 2 题

AIME 2001 II — Problem 2

专题: Contest Math
难度: L4
来源: AIME

题目详情

Problem

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 $M-m$ .

解析

Solution

Let $S$ be the percent of people who study Spanish, $F$ be the number of people who study French, and let $S \cap F$ be the number of students who study both. Then $\left\lceil 80\% \cdot 2001 \right\rceil = 1601 \le S \le \left\lfloor 85\% \cdot 2001 \right\rfloor = 1700$ , and $\left\lceil 30\% \cdot 2001 \right\rceil = 601 \le F \le \left\lfloor 40\% \cdot 2001 \right\rfloor = 800$ . By the Principle of Inclusion-Exclusion,

$S+F- S \cap F = S \cup F = 2001$ For $m = S \cap F$ to be smallest, $S$ and $F$ must be minimized.

$1601 + 601 - m = 2001 \Longrightarrow m = 201$ For $M = S \cap F$ to be largest, $S$ and $F$ must be maximized.

$1700 + 800 - M = 2001 \Longrightarrow M = 499$ Therefore, the answer is $M - m = 499 - 201 = \boxed{298}$ .