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 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 :

XN + YN + ZN = 2008

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

 No Solution Yet Submitted by K Sengupta No Rating

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 computer solution | Comment 1 of 3

DEFDBL A-Z
FOR n = 2 TO 99999
FOR x = 1 TO 2008 ^ (1 / n) / 3
FOR y = x TO 2008 ^ (1 / n) / 2
zn = 2008 - INT(x ^ n + .5) - INT(y ^ n + .5)
z = INT(zn ^ (1 / n) + .5)
IF INT(z ^ n + .5) = zn THEN
PRINT n, x; y; z, x ^ n; y ^ n; z ^ n
END IF
NEXT
NEXT
NEXT

finds

` N             X  Y   Z      X^N  Y^N  Z^N 2             6  6  44      36  36  1936 2             10  12  42    100  144  1764 3             4  6  12      64  216  1728`

 Posted by Charlie on 2013-11-08 13:34:46

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