Last week, we played around with magic squares. If you haven’t had a chance to try the puzzle yet, click here.
It isn’t possible to make a 2×2 magic square. Each row, column, and long diagonal would have to sum to 5, and if you put the digit 1 in the top left corner, you would need a four in the top right and the bottom left.
You can do a 3×3. Below is an example:
2 7 6
9 5 1
4 3 8
For the last problem, we need to remember the fact from an earlier week that
1 + 2 + 3 + … + n^2 = n^2 ( n^2 + 1)/2.
Or you could have tried to show this again. Then, noting that each row has to have the same sum, we know that the sum in each row, column, and long diagonal must be n(n^2 +1)/2.