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 A quite even partition (Posted on 2005-08-28)
Can you partition the numbers 1, 2, 3, ... nē in n separate subsets, each with n numbers, all subsets having the same sum?

 Submitted by Federico Kereki Rating: 4.3333 (3 votes) Solution: (Hide) Write a matrix like this; notice the diagonal: 1, n+1, 2n+1, ..., and the way the terms "grow" cyclically and to the right:```1 2 3 ... n 2n n+1 n+2 ... 2n-1 3n-1 3n 2n+1 ... 3n-2 ... ... ... ... ... (n-1)n+2 (n-1)n+3 (n-1)n+4 ... (n-1)n+1``` Now subtract 0 from the first row (!!), subtract n from the second row, 2n from the third, and so on, and (n-1)n from the n-th row. You get:```1 2 3 ... n n 1 2 ... n-1 n-1 n 1 ... n-2 ... ... ... ... ... 2 3 4 ... 1``` Each column has a permutation of the numbers 1, 2, ... n, so the sum is the same. Hence, the columns of the original matrix are a solution.

 Subject Author Date re: The Odd-Sided Squares Principle Bruno 2005-09-01 14:06:32 Another half solution McWorter 2005-08-31 18:27:44 re: My solution Ken Haley 2005-08-31 06:00:28 My solution Ken Haley 2005-08-31 05:57:47 re: The Odd-Sided Squares Principle Federico Kereki 2005-08-31 03:21:40 re: The Odd-Sided Squares Principle McWorter 2005-08-30 16:21:31 The Odd-Sided Squares Principle Bruno 2005-08-30 11:49:37 re(2): Calendar Magic McWorter 2005-08-29 16:32:01 re: Calendar Magic Bractals 2005-08-29 16:23:22 re(2): Another (similar) way (To Mc) pcbouhid 2005-08-29 11:56:05 re: Another (similar) way (To Mc) McWorter 2005-08-29 03:23:28 Another (similar) way (To Mc) pcbouhid 2005-08-29 03:06:53 re: Calendar Magic McWorter 2005-08-29 02:50:05 A simpler explanation Gamer 2005-08-29 02:01:09 Calendar Magic McWorter 2005-08-29 00:11:06 A process for creating the subsets Bob Smith 2005-08-28 22:25:42 Solution Bractals 2005-08-28 16:39:14 Half Solution owl 2005-08-28 14:54:58

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