AIME 2010 II · 第 1 题

AIME 2010 II — Problem 1

专题: Contest Math
难度: L4
来源: AIME

题目详情

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 $1000$ .

解析

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, $N = 8640 \equiv \boxed{640} \pmod {1000}$ .

Video Solution

https://www.youtube.com/watch?v=TVlHqIgMEVQ