Pigeonhole in Calendar

Solution

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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 a_i, 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 a_i 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.

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