Show that for any given 52 integers, there exist two of them whose sum, or else whose difference, is divisible by 100.

Proof:

Let be a selected 52 integers. Present each or where. If the ordinal remainder exceeds 50 then increase quotient by 1 to make remainder. For example: if 75 in A, then 75=100*2-25, instead of 75=100*1+75; however, if 125 is in A, then 125=100*1+25.

Let. Set B contains 52 remainders between 0 and 50. By Pigeonhole principle, there are at least two remainders for. The corresponding two numbers could be one of the following and . Therefore, one of the must be divisible by 100.