## Pigeonhole in Calendar

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.

Solution 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.

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. • Six integers out of 10: Pigeonhole Principle
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