Download PDF by Daniel J. Velleman: American Mathematical Monthly, volume 117, May 2010

By Daniel J. Velleman

ISBN-10: 0883851857

ISBN-13: 9780883851852

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Extra resources for American Mathematical Monthly, volume 117, May 2010

Example text

However, because there are at least five cells in the fifth column that contain 4−, an integer must be repeated in that column, which is a contradiction. Incidentally, the same conclusion holds for n ≥ 2 if one allows one of the cells to be empty (hence one column to have only k − 1 distinct integers), and does not permit it to be part of a transversal. In this case, delete a column that does not have the empty cell, and argue as in the preceding proof. 4. THE CASE n ≥ 2k − 1. A moment’s thought shows that it suffices to establish this case when n = 2k − 1.

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American Mathematical Monthly, volume 117, May 2010 by Daniel J. Velleman


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