perplexus.info :: Just Math : A! + B! + C! = 3<sup>D</sup>

A! + B! + C! = 3^D (Posted on 2009-08-27)

Determine all possible quadruplet(s) (A, B, C, D) of nonnegative integer(s), with A < B < C, that satisfy this equation:

A! + B! + C! = 3^D

See The Solution Submitted by K Sengupta

Rating: 5.0000 (1 votes)

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Subject	Author	Date
re: Metaphysics?	Charlie	2009-08-29 00:41:21
Metaphysics?	ed bottemiller	2009-08-28 21:02:36
Solution	Dej Mar	2009-08-28 11:18:06
Spreadsheet & Computer without proof	brianjn	2009-08-27 22:46:37
analytical solution	Daniel	2009-08-27 11:33:12

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