In any set of 181 square integers, prove that one can always find a subset of 19 numbers, the sum of whose elements is divisible by 19.

(In reply to conceptual solution by Cory Taylor)

The fact that they are square integers does insure that they are positive integers, so two integers with the same remainder can't cancel. We'd need pairs with remainders like 17 and 2 to cancel. Of course 19 with the same remainder would bring you back to the start, but then you'd have exhausted your 19.

Posted by Charlie on 2003-03-17 03:58:40