I need to solve a 3x3 magic square -1, 2, -3, -4, 6, -9, -12, 18, -36. I know the answer should be -13... I think. Help?
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To find out the common sum, X, add all the numbers together: -39 and divide by the number of rows/columns (3) to get -13. The reason for this can be seen by labelling the squares A-I. If we add up the rows we know the same sum X must apply, so A+B+C=X, D+E+F=X, G+H+I=X. Add these equations together: A+B+C+D+E+F+G+H+I=3X, so X=(1/3)(sum of all the numbers). Now write down the sequence in order: -36 -12 -9 -4 -3 -1 2 6 18. 

-36+a+b=X=-13, so -36 requires two other numbers a,b such that a+b=23. There are only 3 positive numbers and no pair adds up to 23. So we can have no row, column or diagonal containing -36. This invalidates the magic square. Furthermore, there have to be at least two ways to involve -36 in a sum.

Take the row, column or diagonal containing 18. Two numbers a, b added to 18 must come to -13. So a+b=-31. Again, no pair will give us this sum. This further invalidates the magic square.

The standard magic square cannot be constructed from the given numbers.

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