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The integers 1, 2, ..., 10 are written on a circle, in any order. Show that there are 3 adjacent numbers whose sum is 17 or greater.
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There are 10 triples of adjacent numbers with sums S1, S2, ..., S10. If each is less than 17, they all add up to 16×10 = 160, at most. However, in the latter sum each of the numbers 1, 2, ..., 10 appears 3 times, so that the sum must be 3×55 = 165, at least. (55 = 1 + 2 + ... + 10.) It follows that some of Si are bound to be greater than 16.
Copyright © 1996-2009 Alexander Bogomolny
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