# 2006 AIME I Problems/Problem 2

## Problem

Let set $\mathcal{A}$ be a 90-element subset of $\{1,2,3,\ldots,100\},$ and let $S$ be the sum of the elements of $\mathcal{A}.$ Find the number of possible values of $S.$

## Solution

The smallest $S$ is $1+2+ \ldots +90 = 91 \cdot 45 = 4095$. The largest $S$ is $11+12+ \ldots +100=111\cdot 45=4995$. All numbers between $4095$ and $4995$ are possible values of S, so the number of possible values of S is $4995-4095+1=901$.

Alternatively, for ease of calculation, let set $\mathcal{B}$ be a 10-element subset of $\{1,2,3,\ldots,100\}, and let$T$be the sum of the elements of$\mathcal{B}$. Note that the number of possible$S$is the number of possible$T=5050-S$. The smallest possible$T$is$1+2+ \ldots +10 = 55$and the largest is$91+92+ \ldots + 100 = 955$, so the number of possible values of T, and therefore S, is$955-55+1=\boxed{901}.