Can you partition the numbers 1, 2, 3, ... nē in n separate subsets, each with n numbers, all subsets having the same sum?

Bractals' solution completed.

Arrange the numbers in a "calendar" reading left to right, top to bottom.

  1  2  3    .  .  .  n
n+1 n+2   .  .  .  2n
.
.
.
n(n-1)+1  .  .  .  n^2

Superimpose over this array the "clock numbers" addition table modulo n.

 0   1  2  .  .    .  n-1
 1   2  3  .  .  n-1  0
 2   3  4  . n-1 0   1
 .
 .
 .
n-1 0  1  .  .    .  n-2

Now put in the 0-th subset all those numbers below the clock number 0, put in the 1-th subset all those numbers below the clock number 1, and so on, finally putting in the (n-1)-th subset all those numbers below the clock number n-1.  By calendar magic, all these n subsets have n elements and have the same sum.

Posted by McWorter on 2005-08-29 00:11:06