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Re: X, Y and Z are positive integers, and X+2Y+2Z=13. Z=? [#permalink]
15 May 2012, 22:36

Expert's post

X, Y and Z are positive integers, and X+2Y+2Z=13. Z=?

(1) X>Y>Z. If z=2 then the least values of y and x are 3 and 4 respectively, in this case x+2y+2z=14>13, so z must be less than 2 and since given that it's a positive integer then it can only be 1. Sufficient.

(2) None of them is equal to 4. Not sufficient, consider: x=3, y=2, z=3 and x=3, y=3, z=2.

Re: X, Y and Z are positive integers, and X+2Y+2Z=13. Z=? [#permalink]
27 May 2012, 19:43

(1) X>Y>Z. If z=2 then the least values of y and x are 3 and 4 respectively, in this case x+2y+2z=14>13, so z must be less than 2 and since given that it's a positive integer then it can only be 1. Sufficient.

Bunuel, any reason why you did not consider Z=1 so in this case case y=2 and z=3 and then the solution is insufficient.

Re: X, Y and Z are positive integers, and X+2Y+2Z=13. Z=? [#permalink]
28 May 2012, 02:35

Expert's post

rggoel9 wrote:

(1) X>Y>Z. If z=2 then the least values of y and x are 3 and 4 respectively, in this case x+2y+2z=14>13, so z must be less than 2 and since given that it's a positive integer then it can only be 1. Sufficient.

Bunuel, any reason why you did not consider Z=1 so in this case case y=2 and z=3 and then the solution is insufficient.

I guess you mean z=1, y=2 and x=3 (z<y<z). This values are not valid since they does not satisfy x+2y+2z=13. _________________

Re: X, Y and Z are positive integers, and X+2Y+2Z=13. Z=? [#permalink]
30 May 2012, 05:01

another way to look at the question is that the sum is 13... and it is x+2y+2z, which means that z has to be odd as 2z+2y will be even and hence z has to be 1, bcoz any other odd and the condition x>y>z does not give 13.

Originally posted on MIT Sloan School of Management : We are busy putting the final touches on our application. We plan to have it go live by July 15...