# 2001 AIME II Problems/Problem 2

## 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 $M-m$.

## Solution

Let $S$ be the percent of people who study Spanish, $F$ be the number of people who study French, and let $S \cup F$ 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 $M - m = 499 - 201 = \boxed{298}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 