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Re: If x, y, and z are 3 positive consecutive integers such that x < y < z [#permalink]

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20 Dec 2006, 15:35

mm007 wrote:

If x, y, and z are 3 positive consecutive integers such that x < y < z, what is the remainder when the product of x, y, and z is divided by 8?

(1) (xz)^2 is even

(2) 5 . y^3 is odd

Going for D here ... wait for OA though

From 1) possible cases - (x even, z even) or (x even, z odd) or (x odd, z even)
But from the question stem ...x < y < z , here x and z are both either even or odd.
so only even possible ..that means multiple of 8

From 2) y ^3 has to be odd..so y odd ..so x and z are even...divisible by 8

Re: If x, y, and z are 3 positive consecutive integers such that x < y < z [#permalink]

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20 Dec 2006, 16:54

I would go for D

question gives: x<y<z and x,y,z and consecutive ints.

question asks: remainder if xyz/8 ?

Statement 1 : (XZ)^ 2 is even
------------------------------------
from the question, x and z : either both even or both odd
XZ * XZ is even so XZ must be even --> x and z both are even
therefore, y is odd

since the product xyz consists of two even factors, it must be divisible by 2 and also by 4 --> it is divisible by 8

Statement 1 is sufficient

Statement 2: 5 y^3 is odd
--------------------------------
5 is odd --> y^3 must be odd
so, y also must be odd
if y is odd, then x and z are even integers
for the same reasoning in statement 1, the product xyz is divisible by 8

Re: If x, y, and z are 3 positive consecutive integers such that x < y < z [#permalink]

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30 Oct 2014, 17:51

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