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Sum three, get one (Posted on 2007-04-25) Difficulty: 2 of 5
Determine all possible positive integer solutions of this equation:

n! = a! + b! + c!

  Submitted by K Sengupta    
Rating: 2.5000 (2 votes)
Solution: (Hide)
If possible, let n < = a, b, c
Then, a! + b! + c! > = 3n > n, a contradiction.
Thus, n > a, b, c

If in addition, n > 3, then:
n! > 3 (n-1)! > = a! + b! + c!. This is a contradiction

Accordingly, it follows that: n = 1, 2, or 3.

A check for these values of n, yields:

3! = 2! + 2! + 2!.

Hence n =3; a=b=c =2 is the only possible solution to the given problem.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Solution with explanationFrankM2020-04-09 11:47:30
SolutionSolutionhoodat2007-04-26 08:57:19
Possible solution?Mike C2007-04-25 10:13:54
SolutionOne and only oneOld Original Oskar!2007-04-25 10:06:05
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