2010 AIME II Problems/Problem 1
Problem
Let be the greatest integer multiple of all of whose digits are even and no two of whose digits are the same. Find the remainder when is divided by .
Solution
If an integer is divisible by , it must also be divisible by since is a factor of . It is a well-known fact that, if is divisible by , the sum of the digits of is a multiple of . Hence, if contains all the even digits, the sum of the digits would be , which is not divisible by and thus . The next logical try would be , which happens to be divisible by . Thus, .
See also
2010 AIME II (Problems • Answer Key • Resources) | ||
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