Pigeonhole in Calendar
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Given 14 or more integers from {1, 2, ... , 28} there exist four of the given integers which can be split into two groups of two with the same sum.
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Arrange the 28 integers in four rows of seven integers each as below.
1 2 3 4 5 6 7
8 9 10 11 12 13 14
15 16 17 18 19 20 21
22 23 24 25 26 27 28
The 14 given integers lie in 14 cells of the above 4x7 array. As in the proof of
#26, let ai, i=1,...,7, be the number of pairs of rows of the array which each
contain a given integer in the i-th column of the array. Then the sum of the ai is
at least 7 since the minimum sum occurs when each column has two given integers.
Since the number of pairs of rows of the array is 4(4-1)/2=6<7, some 2 by 2 subarray
consists of given integers. Hence the sum of the two diagonal elements of this subarray equals the sum of its two off-diagonal elements.
Copyright © 1996-2009 Alexander Bogomolny
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