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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?

 See The Solution Submitted by Federico Kereki Rating: 4.3333 (3 votes)

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 re: The Odd-Sided Squares Principle | Comment 13 of 18 |
(In reply to The Odd-Sided Squares Principle by Bruno)

Thanks, Bruno, for introducing an entirely different way to solve this problem.

In answer to your question, the literature on magic squares may have a scheme related to yours.

I have some knowledge of something called orthogonal latin squares, which may be related to your method.  Given an odd integer n, form the latin square L, where the (i,j)-entry of L is the remainder mod n of i+j.  Form the latin square L', where the (i,j)-entry is the remainder mod n of 2i+j.  Then L and L' are orthogonal latin squares.  This means that for distinct ordered pairs (i,j) and (s,t), the ordered pairs (i+j,2i+j) and (s+t,2s+t) are also distinct.

Whenever you have a pair of orthogonal latin squares L and L' of size m (m can be even but not of the form 4k+2) in the integers 0 to m-1, you can form a (near) magic square of size m by placing in the (i,j) position the number L(i,j)m+L'(i,j)+1.  Extra work is required to get the diagonals to work along with the rows and columns.

Here's an example for m=4.

L

0 1 2 3
1 0 3 2
2 3 0 1
3 2 1 0

L'

0 1 2 3
2 3 0 1
3 2 1 0
1 0 3 2

Superimposing L' on L produces the ordered pair array

(0,0) (1,1) (2,2) (3,3)
(1,2) (0,3) (3,0) (2,1)
(2,3) (3,2) (0,1) (1,0)
(3,1) (2,0) (1,3) (0,2)

Regarding these pairs as two digit numbers base 4, we get

0   5 10 15
6   3 12  9
11 14  1  4
13   8  7  2

Adding 1 to each entry gives the near magic square of size 4.

 Posted by McWorter on 2005-08-30 16:21:31

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