Can you partition the numbers 1, 2, 3, ... nē in n separate subsets, each with n numbers, all subsets having the same sum?
I found a solution, and quickly read through all the comments, and didn't see it. It's probably isomorphic to a solution given, but it wasn't obvious to me at first glance. Anyway, here's my solution (using 4x4 as an example, but it works for any n):
Arrange the numbers in an nxn square numbering left to right, then top to bottom:
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
The first subset is the major diagonal of the square--upper left to lower right. (in our example 1,6,11,16).
The second set is the (n-1)-number diagonal to the immediate right of the first, wrapping around to the left column when you hit the right edge (2, 7, 12, 13)
The third set is the next diagonal immediately to the right of that, again wrapping around when you hit the right edge (3, 8, 9, 14).
The fourth set is the 4th diagonal immediately to the right of the previous (4, 5, 10, 15).
That completes the 4x4 case--all subsets add up to the same sum, 34. The same procedure works for any size square. Seems pretty simple.
My solution
