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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)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): No Subject Comment 8 of 8 |
(In reply to re: No Subject by Dej Mar)

I've just taken this a small step further.  Try swapping Rows/Columns 1 with 4 and 2 with 5, but I still have 16 at the centre :-(
  Posted by brianjn on 2008-07-30 04:03:35

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