2010 AIME II Problems/Problem 1
Contents
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, .
Video Solution
https://www.youtube.com/watch?v=TVlHqIgMEVQ
See also
2010 AIME II (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
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