(In reply to
re: solution for n=12 by ed bottemiller)
Ady,
I let my computer run overnight, searching for lower sets than that previously mentioned. What I did was the following, for a set of 12 numbers {a, b, c, d, e, f, g, h, i, j, k, l}:
1.) Set a = 1, as any set starting with >1 can be reduced as mentioned in a previous post
2.) Set the limit I'd like to start my search at (in this case, 72)
3.) Set l = limit, this fixed our 1st and 12th values, to make the looping process less time consuming
4.) Began looping through all values for b  k, beginning at b+1, c+1, .... and ending at our set limit
5.) If a set was found, with all values less than or equal to limit, I displayed that set, and broke out of the loop, reduced limit to limit  1, and restarted the loop
I continued the above process, until a result wasn't found for our given limit. After displaying the solution {1, 2, 3, 8, 13, 23, 38, 41, 55, 64, 68, 72} already proposed, it failed to find a solution for a limit of 71.
So, that means EN = 72, and any set of 12 values < 72 will result in at least one duplicate sum.

Posted by Justin
on 20100720 14:18:14 