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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Solution

| Comment 3 of 5 |

There are four quadruplets (A, B, C, D) of nonnegative integers with A < B < C, that satisfies the equation:
A! + B! + C! = 3^D

(0, 2, 3, 2)
0! + 2! + 3! = 3²
1 + 2 + 6 = 9
(1, 2, 3, 2)
1! + 2! + 3! = 3²
1 + 2 + 6 = 9
(0, 2, 4, 3)
0! + 2! + 4! = 3³
1 + 2 + 24 = 27
(1, 2, 4, 3)
1! + 2! + 4! = 3³
1 + 2 + 24 = 27

Posted by Dej Mar on 2009-08-28 11:18:06

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