# Difference between revisions of "2001 AIME I Problems/Problem 2"

## Problem

A finite set $\mathcal{S}$ of distinct real numbers has the following properties: the mean of $\mathcal{S}\cup\{1\}$ is $13$ less than the mean of $\mathcal{S}$, and the mean of $\mathcal{S}\cup\{2001\}$ is $27$ more than the mean of $\mathcal{S}$. Find the mean of $\mathcal{S}$.

## Solution

Let x be the mean of S. Let a be the number of elements in S. Then, $$\frac{ax+1}{a+1}=x-13$$ and $$\frac{ax+2001}{a+1}=x+27$$ $$\frac{ax+2001}{a+1}-40=\frac{ax+1}{a+1}$$ $$\frac{2000}{a+1}=40$$ so $$2000=40(a+1)$$ $$a=49$$ We plug that into our very first formula, and get: $$\frac{49x+1}{50}=x-13$$ $$49x+1=50x-650$$ $$x=651$$