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 we include all the even digits for the greatest integer multiple, we find that it is impossible for it to be divisible by $3$, therefore by $36$ as well. The next logical try would be $8640$, which happens to be divisible by $36$. Thus $N = 8640 \pmod {1000} = \boxed{640}$

See also

2010 AIME II (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
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All AIME Problems and Solutions

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