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Problem of the Week 1148
Seamless Brickwork
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The diagram below shows a special arrangement of three rows of four bricks each:
In each row, the size is a permutation of (1, 2, 3, 4). Moreover, the arrangement is such that no two of the vertical dividers between bricks is aligned. In other words, each of the 9 partial sums 1, 2, 3, 4, 5, 6, 7, 8, 9 arises once and once only when the sums along each row are formed. Three rows are used because, ignoring the everpresent terminal sum of 10, each row contributes three partial sums, for a total of 9.
Find such an arrangement using 1, 2, 3, 4, 5, 6 in each row. The possible sums are then 1, 2, 3, 4, ..., 20 and each row contributes five, so one needs four rows.
This problem makes sense for any even n = 2k, but I am not aware of the current bestknown results. Does anyone know? Because the total number of sums is n(n + 1)/2 − 1 and each row contributes n − 1 sums, the number of rows is k + 1. Thanks to Rob Pratt using sophisticated compuational methods, I can say that there are solutions for n = 4, 6, 8, 10, 12, 14, 16, 20, 22, 24, 26, 30, 34, 38.
Source: Suggested by Tom Halverson who heard it in a recent lecture by Barry Cipra.
© Copyright 2011 Stan Wagon. Reproduced with permission.
