Determine all possible quadruplets (p, q, r, s) of positive integers with p< = q< = r satisfying the equation p! + q! + r! = 2^s.
(In reply to
computer findings by Charlie)
Wow! I like that program, it's very short and sweet; what language is that? Is it Python? I've heard good things about Python.
My brief paper and pencil calculations gave me the same 4 possibilities as those found by Charlie's program - of course, I didn't go as far as p+q+r = 512. I made a quick table of powers of 2 up to 2^16 = 65536, and factorials up to 10! = 3 628 800. Then I just looked to see how I could add up three of the factorials to get one of the powers, starting from the smallest cases and working down the table.
However, neither Charlie nor I have shown that there could not exist a much larger solution somewhere... who's to say that we might not find a solution where each term in the sum has 100 digits, perhaps? Or even more?
I just mentioned this problem to a friend of mine and he had a great idea of FACTORING the left side of the equation - I think that approach could resolve this problem. I will think some more on it. Good question!
-John
re: computer findings
