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More or less magic square (Posted on 2004-06-22) Difficulty: 3 of 5
Fill a 3x3 square with the numbers from 1 to 9, so the four sides and the two diagonals share the same sum, which is to be as large as possible.

See The Solution Submitted by Federico Kereki    
Rating: 3.2500 (4 votes)

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Solution Puzzle Resolution Comment 11 of 11 |

The largest available digits are 6, 7, 8 and 9; and so these numbers are placed at the four corners as follows:

8 _ 9
_ _ _
6 _ 7


After a little trial and error, it is observed that the largest possible sum with a solution occurs at 18.

Thus, value at cell 21 = 18-8-6 =12; cell 23 = 18- 16=2;
cell 12 = 18 - 17 = 1; cell 32 = 18-6-7= 5. The structure at this
point are as follows:

8 1 9
4 _ 2
6 5 7

The only remaining number is 3, which will correspond to cell 22.

Consequently, the required square as completed is:

8 1 9
4 3 2
6 5 7


  Posted by K Sengupta on 2007-06-19 10:59:54
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