If I have a set of 10 integers, is it possible to prove there are two that the difference is by a multiple of nine?
My instinct says you can find two that differ by a multiple of 5 but not 9
2026-03-28 22:37:00.1774737420
Pigeon Hole theory with 10 ints
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Hint: If you divide a number by $9$, there are nine possible remainders. Thus, in any set of ten numbers, there must be two with the same remainder when divided by $9$.