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In a room there are 10 people, none of whom are older than 100 (ages are given in whole numbers only) but each of whom is at least 1 year old. Prove that one can always find two groups of people (possibly intersection, but different) the sums of whose ages are the same.

I know we have to use the pigeon hole principle. But I don't know how to find the pigeons and pigeonholes. Can someone help me out?

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    It seems as though if you ca$n$ solve it with possible intersection, you can solve it without intersection: If you remove the same person from two sets whose sum of ages are the same, the sum of ages stays the same.2012-12-09

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Hint: what is the maximum sum of all the ages in the room? How many subsets are there of people in the room?

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    OH! So there are 1001 pigeonholes and 1024 pigeons! Got it! Thank you Brian!2012-12-09