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# If Y is a factor of Z , is X even? (1) \frac{X^{3}Z}{Y} + X

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Manager
Joined: 14 May 2009
Posts: 189
If Y is a factor of Z , is X even? (1) \frac{X^{3}Z}{Y} + X [#permalink]

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03 Jun 2009, 00:24
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If $$Y$$ is a factor of $$Z$$, is $$X$$ even?

(1) $$\frac{X^{3}Z}{Y} + X$$ is even

(2) $$XY$$ is even

--== Message from GMAT Club Team ==--

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Senior Manager
Joined: 15 Jan 2008
Posts: 267

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03 Jun 2009, 01:30

1, insufficient

as Y is a factor of Z, we know z/y = K, but k can be even or odd..

x(x^2 K + 1) is even.. this is possible with both X is even and odd.
hence inconclusive.

2, talks abt X and Y.. no conclusive infomration abt X.

combined.. they dont give enough informatin abt X.. hence conclusive..
Intern
Joined: 27 May 2009
Posts: 6

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03 Jun 2009, 07:11

if y is a factor of z, then z = yK, where k is some integer constant.

1. x^3.z/y + x can be written as x^3.yk + x. Both sides of + sign can be EVEN or ODD -> insufficient

2. xy is even. X can odd or even, if y is even -> not sufficient

However, when we combine 1 and 2

x^3.yk + x can be written as x^2.xy.k, which is even since, from condition 2, xy is even.

even + x = even

hence x has to even.

VP
Joined: 28 Dec 2005
Posts: 1499

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03 Jun 2009, 09:28
I think E for this one.

From stat 2, you know that either x or y can be even. insuff.

From stat 1, you can deduce that both terms have to be either both odd or both even to get an even sum. So case 1 is Z/Y is even, which means that for the sum to be even, X (and therefore X^3) are even as well. Case 2 is Z/Y being odd, then X (and therefore X^3) must be odd as well to make the sum even. X can be even or odd, so insuff.

Together, you know X can be even or odd, as can Y ... if Y is even, then Y/Z can be even or odd (ie 10/2) or even (12/2). Depending on what the quotient is , X can be even or odd
Manager
Joined: 14 May 2009
Posts: 189

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03 Jun 2009, 15:46
1
KUDOS
If $$Y$$ is a factor of $$Z$$, is $$X$$ even?

(1) $$\frac{X^{3}Z}{Y} + X$$ is even

(2) $$XY$$ is even

Question:$$(X$$ even$$)$$?

(1) $$\frac{X^{3}Z}{Y} + X$$ is even

$$\longrightarrow X(\frac{X^{2}Z}{Y} + 1)$$ is even

$$\longrightarrow X$$ even OR $$\frac{X^{2}Z}{Y} + 1$$ even

If $$X$$ even then we have a YES, but if $$X$$ not even then:

$$\longrightarrow \frac{X^{2}Z}{Y} + 1$$ even

$$\longrightarrow \frac{X^{2}Z}{Y}$$ odd

$$\longrightarrow \frac{XZ}{Y}$$ odd

$$\longrightarrow X odd & \frac{Z}{Y}$$ odd

(2) $$XY$$ even
$$\longrightarrow (X$$ even, $$Y$$ odd) or ($$Y$$ even, $$X$$ odd) or ($$X & Y$$ even)

So we have a YES case & a NO case $$\longrightarrow INSUFFICIENT$$

(1&2) Combining

Look at (1) and look at the case $$X odd$$. From (2) we know that $$Y even$$. But (1) can still be true because $$Z=MY$$ where $$M odd$$ and $$Y even$$, and everything holds.

NO case (X odd)
ie $$X=1$$, $$Y=2$$, $$Z=6$$.
$$XY = 2$$ (even-- 2 is satsified)
$$\frac{X^{3}Z}{Y} + X = \frac{1^{3}6}{2} + 1 = \frac{6}{2} + 1 = 3 + 1 = 4$$ (even)

YES case (X even)
ie $$X=2$$, $$Y=2$$, $$Z=6$$.
$$XY = 4$$ (even-- 2 is satsified)
$$\frac{X^{3}Z}{Y} + X = \frac{2^{3}6}{2} + 2 = \frac{48}{2} + 2 = 24 + 2 = 26$$ (even)

YES & NO$$\longrightarrow INSUFFICIENT$$

$$\longrightarrow INSUFFICIENT$$

Final Answer, $$E$$.
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Senior Manager
Joined: 16 Jan 2009
Posts: 349
Concentration: Technology, Marketing
GMAT 1: 700 Q50 V34
GPA: 3
WE: Sales (Telecommunications)

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03 Jun 2009, 16:13
Hello mjGMAT,

if
mjGMAT wrote:
if y is a factor of z, then z = yK, where k is some integer constant

z/y=k I think you have used z/y = yk
mjGMAT wrote:
x^3.yk + x can be written as x^2.xy.k, which is even since, from condition 2, xy is even.

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Lahoosaher

Senior Manager
Joined: 16 Jan 2009
Posts: 349
Concentration: Technology, Marketing
GMAT 1: 700 Q50 V34
GPA: 3
WE: Sales (Telecommunications)

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03 Jun 2009, 16:16
+1

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If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.

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Lahoosaher

Re: Tough DS 6   [#permalink] 03 Jun 2009, 16:16
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