There are 40 ways to make sums of three distinct positive integers total 25. (1+2+22 is such a sum, but 1+12+12 and 1+2+3+19 are not.)
How many different ways can three distinct positive integers sum to 1000?
(In reply to
re: Curious second differences by Federico Kereki)
Following that course of thought, I found an interesting formula. If N=6K+L (in the original problem, N=1,000) then the number of different ways is 3K²+(L-3)K -- plus 1, if L=0.
For N=1,000, K=166 and L=4, and the formula correctly produces 82,834.
|
Posted by Oskar
on 2004-07-22 13:04:49 |