Question 1096931: Use numbers 3 to 11 to fill in a 3x3 magic square ?
Answer by greenestamps(13198)   (Show Source):
You can put this solution on YOUR website!
In a 3x3 magic square, there are 8 places where there are 3 squares in a straight line: 3 rows, 3 columns, and 2 diagonals.
Each square in a corner is in 3 of those - one row, one column, and one diagonal.
Each square on a side is in only 2 of those - one row and one column.
And the square in the middle is in 4 of those - one row, one column, and both diagonals.
When you are filling in a 3x3 magic square using consecutive integers, the middle number always goes in the middle square; and the sum of each row, column, and diagonal is 3 times that number. So when using the numbers 3 to 11, the number 7 is in the middle, and the magic sum is 21.
(1) Make a list of all the ways of adding three of the numbers 3 to 11 to get a sum of 21:
3+7+11
3+8+10
4+6+11
4+7+10
4+8+9
5+6+10
5+7+9
6+7+8
(2) The numbers 3, 5, 9, and 11 each appear twice in that list of sums, so they must be the numbers on the sides of the square. Place them any way you want so that the 3 and 11 are in a straight line and the 5 and 9 are in a straight line:
The numbers yet to be placed are 4, 6, 8, and 10. Find where each goes by logical analysis.
In the first column, the top and bottom numbers have to add to 18, so they have to be 8 and 10. And in the first row, the first and last numbers have to add to 16, so they must be 10 and 6.
But that means the top left square must have the number 10:
And then it is easy to see where to place the 4, 6, and 8 to make all the rows, columns, and diagonals have the magic sum of 21:
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