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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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Wrap didn't work | Comment 1 of 8
For magic squares where n is odd and the numbers are consecutive one can fill the square beginning at the central top square and follow a diagonal wraparound route (there are things to contend with when one engages the edges of the square or the diagonal route is interrupted, but I leave that to the reader to follow-up) to completion.

I used this process initially beginning with "0" in the central top square and after every multiple of 4 I added another "0".  While it had the desired effect of placing the "0's" the few of the respective sums worked.

In a second attempt I began with a "1" in the central top square and again added "0" after multiples of 4 with similar results.  Back to the drawing board.

....... but then one should expect the second result to be similar because of the cyclic wrapping.  I do note however that in both case one major diagonal was 42 while the other changed.

Edited on July 26, 2008, 9:31 pm
  Posted by brianjn on 2008-07-26 21:18:41

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