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Sum Three Factorials, Get Exponent (Posted on 2007-07-03) Difficulty: 3 of 5
Determine all possible quadruplets (p, q, r, s) of positive integers with p< = q< = r satisfying the equation p! + q! + r! = 2^s.

  Submitted by K Sengupta    
Rating: 4.0000 (1 votes)
Solution: (Hide)
(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.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionsolutionPraneeth Yalavarthi2007-07-06 12:30:17
SolutionSolutionDej Mar2007-07-04 03:30:52
SolutionSolutionhoodat2007-07-03 14:21:40
SolutionFull solutionJer2007-07-03 12:05:30
re(2): computer findingsCharlie2007-07-03 11:57:36
re: computer findingsJohn Reid2007-07-03 11:48:05
Some Thoughtscomputer findingsCharlie2007-07-03 10:31:57
Some ThoughtsJust for startersFederico Kereki2007-07-03 10:26:39
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