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Well i will try to do my best on this one ... am still learning from you guys.specially both of you guys kevin and dahiya.

From the question stem x,y,z are all either +ve or -ve ( intigers or fractions)

from one... the square covers the signe of xyz and x,y,z are intigers and ... insuff from 2 we do not have enough information about the signes of x,y,z ( specially that they could be fractions

Taking one and two ( from one) x,y,z are intigers and from question stem ..the least possible absolute values for x,y,z respectively is 6,10,15

thus sure that they are not all +ve Thus and putting into consideration the question stem they are all -ve... suff thus my answer is c

Well I think that because (xyz)^2 is intiger therfore x,y,z are either intigers or a mix between intigers and complex numbers ( 7/2 and fractions ie : or their product might be 1 or a square root of an intiger ).... to keep the proportion of 5:3:2........

Last edited by yezz on 02 Sep 2006, 05:38, edited 1 time in total.

Note that y=5x/3 and z=5x/2 and that (x,y,z)=(0,0,0) is a solution that satisfies both conditions. If x=y=z=0, xyz is not positive. So the question is, could xyz be positive?

(1) implies that (25x^3/6)^2= 5^4*x^6/3^2*2^2 is an integer

Thus it could be that |x|=6^(1/3)*k^(1/6) where k is an integer

If k=0, x=0 and xyx=0 , but if k=1, x=6^1/3>0. Thus y and z would be positive and xyz would be positive, too. NOT SUFF

(2) 2x+5x+5x/2=19x/2<50 => x<100/19 NOT SUFF

(1) and (2) Both possibilities for x cited in (1) satisfy (2) NOT SUFF