Problem:
The integer is the smallest positive multiple of such that every digit of is either or . Compute .
Solution:
Note that is a common multiple of and . As a multiple of must end in , as the digit is not allowed. As a multiple of must contain a number of 's equal to a multiple of . Hence, in view of the minimality requirement, , and is the answer to the problem.
The problems on this page are the property of the MAA's American Mathematics Competitions