Sum Three Factorials, Get Exponent (Posted on 2007-07-03)

	Submitted by K Sengupta
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(p, q, r, s)= (1, 1, 2, 2); (1, 1, 3, 3), (2, 3, 4, 5), (2, 3, 5, 7} EXPLANATION: If p> =3, then it follows that p! + q! + r! would necessarily possess a factor of 3. This is a contradiction. Accordingly p = 1, 2 Case 1: p=1 If q>=2, then p!+ q! + r! would be odd. This is a contradiction. Hence q=1. If r> =4, then (p!+q!+r!) Mod 4 = 2, which is a contradiction Accordingly we must have r = 2, 3. Now, (p, q, r) = (1,1, 2) gives s = 2; while (p, q, r) = (1,1, 3) gives s = 3. Case 2: p =2 If q> =4, then (p! + q! + r!) Mod 4 = 2, which is a contradiction. Hence q = 2, 3. If q =2, then for similar reasons as before r < 4. It is easily verified that r cannot be 2 or 3. Now if c > = 4, then p! + q! + r! = 4 + r! = 4 mod 8. Hence, q = 2 is a contradiction. Suppose q = 3. Then, we have r! + 8 = 2^s. It is easily verified that c = 3 fails, and c = 4, 5, yields respectively s = 5, 7 If c > = 6, then 16 divides c!, so a! + b! + c! = 8 mod 16, which is a contradiction. Thus, (p, q, r, s)= (1, 1, 2, 2); (1, 1, 3, 3), (2, 3, 4, 5), (2, 3, 5, 7} are the only possible solutions to the given problem.

	Subject	Author	Date
	solution	Praneeth Yalavarthi	2007-07-06 12:30:17
	Solution	Dej Mar	2007-07-04 03:30:52
	Solution	hoodat	2007-07-03 14:21:40
	Full solution	Jer	2007-07-03 12:05:30
	re(2): computer findings	Charlie	2007-07-03 11:57:36
	re: computer findings	John Reid	2007-07-03 11:48:05
	computer findings	Charlie	2007-07-03 10:31:57
	Just for starters	Federico Kereki	2007-07-03 10:26:39