perplexus.info :: Just Math : Some Powers Sum 2008

Some Powers Sum 2008 (Posted on 2013-11-08)

N is an integer ≥ 2 and each of X, Y and Z is a positive integer with X ≤ Y ≤ Z

Determine all possible solutions to this equation :

X^N + Y^N + Z^N = 2008

Extra Challenge: Solve this puzzle without using a computer program.

See The Solution Submitted by K Sengupta

Rating: 5.0000 (1 votes)

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| Comment 2 of 3 |

2008 = 44^2 + 6^2 + 6^2

2008 = 42^2 + 12^2 + 10^2

2008 = 36^2 + 26^2 + 6^2

2008 = 30^2 + 28^2 + 18^2

2008 = 2^3 + 10^3 + 10^3 = 4^3 + 6^3 + 12^3

Posted by Benny on 2013-11-08 17:21:00

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