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:
Because and , it suffices to verify divisibility by and by separately. Because divides , the sum of the digits of is divisible by . Because the digits of are even, their sum must be divisible by , and hence the set of possible digits of is . The maximum value of formed by these digits is , which is divisible by . Thus , and the requested remainder is .
The problems on this page are the property of the MAA's American Mathematics Competitions