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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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re: No Subject | Comment 7 of 8 |
(In reply to No Subject by brianjn)

Swapping columns two and four and swapping rows two and four will provide a solution where a solution exists as it doesn't change the values in a column, row or diagonal. Thus the following two arrangements for holes can be considered the same:

_ X X X X           _ X X X X
X X _ X X           X X X _ X
X X X X _   and   X X X X _
X _ X X X           X X _ X X  
X X X _ X           X _ X X X

There, then, appears to be only three general arrangements.

If the cells in the opposite corners totals are equal, then the other two general arrangements can also be considered the same as swapping either the first row and fifth row or first column and fifth column will not change the totals in columns, rows or diagonals:

X _ X X X           X _ X X X
_ X X X X           X X X X _
X X _ X X   and   X X _ X X
X X X X _           _ X X X X  
X X X _ X           X X X _ X

Edited on July 30, 2008, 8:56 am
  Posted by Dej Mar on 2008-07-30 03:08:55

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