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 $N = 8640 mod 1000 = \fbox{640}$

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2010 AIME II Problems/Problem 1

Problem

Solution