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

(In reply to Calendar Magic by McWorter)

For those unfamiliar with calendar magic here's a proof that all subsets have the same sum.

The number in the i-th row j-th column of the array is (i-1)n+j.  In the addition table for the residues modulo n, each residue occurs exactly once in each row and column of the table.  Hence the formulas for the numbers below a fixed residue in the superposition of the table over the array contain each i, from 0 to n-1 exactly once and contain each j, from 1 to n exactly once.  Hence the sum of all these numbers below a fixed residue sum to

((n-1)n/2)n+(n(n+1)/2=n(n^2+1)/2.

Gamer's simpler explanation (as well as my less simple one) also includes starting with an arbitrary n by n latin square in the numbers 1 to n.  So there are an incredible number of different solutions to this problem.

Posted by McWorter on 2005-08-29 02:50:05