# 2013 USAMO Problems/Problem 5

Given postive integers and , prove that there is a positive integer such that the numbers and have the same number of occurrences of each non-zero digit when written in base ten. The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

## Solution

This solution is adopted from the official solution. Both the problem and the solution were suggested by Richard Stong.

Without Loss of Generality, suppose . By prime factorization of , we can find a positive integer such that where is relatively prime to . If a positive is larger than , then , where is always relatively prime to .

Choose a large enough so that is larger than . We can find an integer such that is divisible by , and also larger than . For example, let and use Euler's theorem. Now, let , and . We claim that is the desired number.

Indeed, since both and are less than , we see that the decimal expansion of both the fraction and are repeated in -digit. And we also see that , therefore the two repeated -digit expansions are cyclic shift of one another.

This proves that and have the same number of occurrences of non-zero digits. Furthermore, also have the same number of occurrences of non-zero digits with .