Determine all possible triplets of integers (n,m,k) satisfying 1!+2!+3!+...+n!=m^k, where n, m and k are greater than 1.

one solution set would be (3,3,2)

there is no other solution till n=8

Edited on February 1, 2007, 4:11 pm

Posted by Vishal Gupta on 2007-02-01 15:39:07