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)

I assume this is what they are saying. Put 1 through n in a simple magic square: (For example if n=5)

12345
23451
34512
45123
51234

Then add 0 to the first row, n to the second row, 2n to the third row...

01 02 03 04 05
07 08 09 10 06
13 14 15 11 12
19 20 16 17 18
25 21 22 23 24

There you have your combinations, reading down. This works since the original arrangement had equal columns, and adding the same number to each of the columns keeps them at the same total.

Posted by Gamer on 2005-08-29 02:01:09