Using the method of mathematical induction

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SUMMARY

The discussion centers on proving that a ten-element set of positive integers, H, ranging from 1 to 99, contains two disjoint subsets A and B with equal sums. The proof utilizes the pigeonhole principle, noting that there are 1022 non-empty proper subsets of H, while the possible sums of these subsets range from 1 to 1000. Since 1022 exceeds the number of distinct sums, it follows that at least two subsets must yield the same sum, confirming the existence of disjoint subsets A and B.

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1. Homework Statement
Let H be a ten- element set of potive integers ranged from 1 to 99. Prove that H has two disjoint subsets A and B so that the sum of the elements of A is equal to the sum of the elements of B.
 
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I think this is most naturally a pigeonhole problem. Pigeonhole hint: there are 2^10 - 2 = 1022 non-empty proper subsets of H, the sum of the elements of any such subset lies in 1..1000 so can take on at most 1000 different values (in fact it's quite less than this, but this is a strong enough statement) which is less than 1022.
 

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