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Use the Pigeonhole Principle (also called the Dirichlet Principle). First, pick 50 different positive integers, and let them be a1, a2, ..., a50, with a1 < a2 < ... < a50. Call this set A. If any of these 50 integers differ by exactly 50, then you're done; so assume that they don't. Then a1+/- 50, a2+/- 50, ..., a50 +/- 50 (using the + sign for those values less than or equal to 50, and the minus sign for those values that are greater than 50) must be the other 50 numbers in the range from 1 to 100. Call this set B. So sets A and B combined represent all the integers from 1 to 100. Now, pick the 51st number. It must come from set B (since it's not one of the first 50 integers picked, which got put in set A). But then it must be equal to ai +/- 50 for some value of i, and hence differs from one of the other numbers chosen by exactly 50.

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Q: How do you prove that among any set of 51 different positive integers whose values are at most 100 there is always a pair of numbers that differ by exactly 50?
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