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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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Solution One way | Comment 2 of 8 |
Rows, columns and major diagonals all total 42.
I have used "0's" for the 'holes' as they were compliant with my spreadsheet grid.

 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

In my prior comment I referred to a wraparound approach.   That is not evident here;  I wonder  if I might interchange some columns and/or rows without destroying the diagonal totals.

I can preserve the row and column totals by simply swapping cols 1 & 5 but the diagonals become 31 and 53.  Mm! With that arrangement I reinstate the diagonal totals if I now exchange row 1 with row 5!


  Posted by brianjn on 2008-07-27 21:41:13
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