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

The total sum of all the integers from 1 to n^2 is (n^2)(n^2+1)/2, so the sum of each subset would have to be (n/2)(n^2+1).
IF n IS EVEN, I can achieve this by using the old trick of adding the smallest to the largest n/2 times, and working my way through the sequence n times.
For example, my first subset would have n^2 and 1, n^2-1 and 2, and so on up to n^2-n/2+1 and n/2. My next subset would continue form there; n^2-n/2 and n/2+1 up to the pair n^2-n+1 and n. Since each subset has n/2 pairs, each pair adding to n^2+1, I get n subsets of n elements, eavh totaling (n/2)(n^2+1).

I don't see readily what to do with n odd, and the yard (and wife) are calling for the lawnmower to be pushed (and I am the designated lawnmower pusher :-) )

Posted by owl on 2005-08-28 14:54:58