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A! + B! + C! = 3D (Posted on 2009-08-27) Difficulty: 2 of 5
Determine all possible quadruplet(s) (A, B, C, D) of nonnegative integer(s), with A < B < C, that satisfy this equation:

A! + B! + C! = 3D

See The Solution Submitted by K Sengupta    
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re: Metaphysics? Comment 5 of 5 |
(In reply to Metaphysics? by ed bottemiller)

Well, for one thing, n!=(n-1)! * n, and doing the algebra, (n-1)! = n! / n, so that 0!=1!/1 = 1. If 0! were 0, then every factorial thereafter would be zero. It's also related to the Gamma function.
  Posted by Charlie on 2009-08-29 00:41:21

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