Question

Any 52 integers two of them who sum or else whose difference is divisible by 100

Answer 1

You have 100 boxes and 52 integers. Pigeon hole does not apply with the integers being the pigeons...

For a proof which keeps the number of pigeons the same:

Consider the "boxes"

(0,100), (1,99), (2,98),…,(50,50)

There are 52 pigeons (the remainders of the integers modulo 100) and 51 boxes.

At least one box must have two integers. If both are y, then their difference is divisible by 100. If one is y and the other 100−y, then their sum is divisible.

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