7th grader wanting to find solution to magic square:

place the integers from -5 to +10 in the magic square so that the total of each row, column, and diagonal is 10. The magic square is 4 squares x 4 squares.

Also 5 x 5 magic square using the integers from -9 to +15. The total of each row, column, and diagonal should be 15.

Thanks

Hi,

I'll show you a method to construct a 3x3 magic square that should work for 4x4 and 5x5 squares. I will place the numbers from 1 to 9 in the square and the sum of each row, column and diagonal will be 15.

The procedure has three steps.

First the rows.

Put the numbers from 1 to 9 into three rows so that the sum of each row is 15. I got

9 1 5
8 4 3
7 6 2
Next the columns.

The row sums are all 15 but the column sums are not. I can, however, move the numbers in any row, as long as they stay in the same row, and the row sums will remain at 15. Move the 4 in the second row to the first column and the 2 in the third row to the first column.

9 1 5
4 8 3
2 7 6

Now the first column sums to 15. Interchange the 6 and 7 in the last row.

9 1 5
4 8 3
2 6 7

Now the row sums and column sums are all 15. Finally the diagonals.

At this point I can interchange any two rows or interchange any two columns and the row sums and column sums with remain unchanged. Interchange the first two rows.

4 8 3
9 1 5
2 6 7

Interchange the second and third columns.

4 3 8
9 5 1
2 7 6

If you want a 3x3 magic square constructed from the numbers from -2 to 6 with the row, column and diagonal sums of 6 just subtract 3 from each of the numbers in the magic square above.

1  0  5
 6  2 -2
-1  4  3

Cheers,
Penny
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