# Difference between revisions of "2010 AIME II Problems/Problem 1"

## Problem

Let $N$ be the greatest integer multiple of $36$ all of whose digits are even and no two of whose digits are the same. Find the remainder when $N$ is divided by $1000$.

## Solution

If an integer is divisible by $36$, it must also be divisible by $9$ since $9$ is a factor of $36$. It is a well-known fact that, if $N$ is divisible by $9$, the sum of the digits of $N$ is a multiple of $9$. Hence, if $N$ contains all the even digits, the sum of the digits would be $0 + 2 + 4 + 6 + 8 = 20$, which is not divisible by $9$ and thus $36$. The next logical try would be $8640$, which happens to be divisible by $36$. Thus, $N = 8640 \equiv \boxed{640} \pmod {1000}$.