Determine all possible quadruplet(s) (A, B, C, D) of nonnegative integer(s), with A < B < C, that satisfy this equation:

A! + B! + C! = 3^D

The explanations why 0! = 1 seem arcane (a collection of "no things" is not a collection of "nothing" etc.etc.). If 0!=zero, then a solution would be A=0, B=1, C=2, D=1 -- but only on every other Thursday... Nothing gained, nothing lost.  A whole weekend to unscrew the inscrutable!

 

Posted by ed bottemiller on 2009-08-28 21:02:36