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.

Submitted by K Sengupta

Rating: 5.0000 (1 votes)

Solution: (Hide)

X Y Z X^N Y^N Z^N 2 6 6 44 36 36 1936 2 6 26 36 36 676 1296 2 10 12 42 100 144 1764 2 18 28 30 324 784 900 3 2 10 10 8 1000 1000 3 4 6 12 64 216 1728
For an explanation, refer to the solution submitted by Charlie in this location.

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
re: more results	Charlie	2013-11-08 17:39:58
more results	Benny	2013-11-08 17:21:00
computer solution	Charlie	2013-11-08 13:34:46

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