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Holes in a magic square (Posted on 2008-07-26) Difficulty: 2 of 5
We know that using the numbers from 1 to 25 (once each), we can build a magic square of order 5, being 65 the magic constant.

Your task is to build a magic square of order 5, using only the numbers from 1 to 20 (once each), leaving one cell empty in each row, in each column, in each main diagonal.

Obviously, the magic constant will be [(1 + 20)/2]*20/5 = 42.

Note: This type of magic square has a name.

See The Solution Submitted by pcbouhid    
Rating: 3.0000 (1 votes)

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No Subject | Comment 6 of 8 |

 These are four solutions which I derived.  They are essentially the one solution with columns and rows interchanged.


I haven't attempted to find a solution where the hole is the central cell and I haven't checked which of the hole patterns match with Dej Mar's arrays.


 0 10 17 14  1
 3 12   0  7 20
 2   9 16 15  0
19  0   4  6 13
18 11  5  0   8

  8 11  5   0 18
20 12  0   7   3
  0  9 16 15   2
13  0   4   6 19
  1 10 17 14  0

  6 19  4 13  0
14   0 17  1 10
15   2 16  0   9
  0 18   5  8 11
  7   3  0 20 12
 

18  0  5  11 8
  3  7  0 12 20
 2 15 16  9  0
19  6  4   0 13
 0 14 17 10  1



I wrote this in MSWord, copied and pasted and brought in some annoying tags [ <o:p></o:p> ] which I've removed.

Edited on July 30, 2008, 3:35 am
  Posted by brianjn on 2008-07-29 22:25:13

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