Determine all possible quadruplets (p, q, r, s) of positive integers with p< = q< = r satisfying the equation p! + q! + r! = 2^s.

2^s must be 4, 8, 32 or 128 as the value is to be equal to the sum of three positive integer factorials.  It may be possible but highly improbable that s may be greater than 7.  Both 0! and 1! produce the value 1 but 0 is not a positive integer, therefore the only probable solutions are:

(1,1,2,2)
(1,1,3,3)
(2,3,4,5)
(2,3,5,7)

Posted by Dej Mar on 2007-07-04 03:30:52