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If Z is a factor of Y , XZ even? (1) The multiple of any two

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Manager
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If Z is a factor of Y , XZ even? (1) The multiple of any two [#permalink]

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New post 03 Jun 2009, 00:32
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If \(Z\) is a factor of \(Y\), \(XZ\) even?

(1) The multiple of any two factors of \(X\) or \(Y\) is odd.

(2) \(\frac{XY}{Z}\) is odd.
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Hades

Kudos [?]: 48 [0], given: 1

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Re: Tough DS 7 [#permalink]

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New post 03 Jun 2009, 01:24
answer shud be C

1, insufficient.. doesnt say anthing abt Z, just says taht X and Y are odd numbers..
2, insuffficient..

XY is even and Z is odd... XZ depends on X .. can be odd or even..
XY is odd and Z is is odd.. XZ is odd ..

Combined..

X and Y are odd numbers... and Z is odd number.. hence XZ is odd..

Kudos [?]: 49 [0], given: 3

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Re: Tough DS 7 [#permalink]

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New post 03 Jun 2009, 08:20
Really ? I got C as well

With statement 2, Y/Z perfectly divide each other, and for the result when multiplied by X to be odd, X must be odd as well. But I dont know anything about Z ... Y/Z is odd and that could happen if Z is even (i.e. 6/2 = 3, or if Z is odd 9/3=3)

From statement 1, we can conclude that X and Y are odd, but nothing about Z

Together, X is odd, Y is odd, and since Y/Z=odd (from statement 2), I concluded that Z is odd as well. Hmmm....

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Manager
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Kudos [?]: 48 [0], given: 1

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Re: Tough DS 7 [#permalink]

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New post 03 Jun 2009, 16:05
Hades wrote:
If \(Z\) is a factor of \(Y\), \(XZ\) even?

(1) The multiple of any two factors of \(X\) or \(Y\) is odd.

(2) \(\frac{XY}{Z}\) is odd.


Question:\((XZ\) even?\()\)

(1) The multiple of any two factors of \(X\) or \(Y\) is odd.

\(\longrightarrow\) X & Y are odd

\(Y\) odd \(\longrightarrow Z\) odd

\(XZ = ODD*ODD = ODD\)

\(\longrightarrow\) SUFFICIENT

(2) \(\frac{XY}{Z}\)

\(\longrightarrow X\) odd & \(\frac{Y}{Z}\) odd

But \(Y=MZ\) where \(M\) is any integer

So \(\frac{Y}{Z}\) odd \(\longrightarrow\) \(\frac{MZ}{Z}\) odd

\(\longrightarrow M\) odd

And doesn't tell you anything about \(Z\), hence \(Z\) can be ODD or EVEN.

Insufficient.

Final Answer, \(A\).
_________________

Hades

Kudos [?]: 48 [0], given: 1

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Re: Tough DS 7 [#permalink]

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New post 03 Jun 2009, 22:58
Agree, the answer is A...

I missed out on a simple statement that..

when X is a factor of Y, and when Y is odd, then X must be ODD...

good one..

Kudos [?]: 49 [0], given: 3

Re: Tough DS 7   [#permalink] 03 Jun 2009, 22:58
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