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

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New post 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
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Hades

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Re: Tough DS 6 [#permalink]

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New post 03 Jun 2009, 01:30
E shud be answer..

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..

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Re: Tough DS 6 [#permalink]

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New post 03 Jun 2009, 07:11
Answer is C

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.

Answer is c.

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Re: Tough DS 6 [#permalink]

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New post 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

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Re: Tough DS 6 [#permalink]

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New post 03 Jun 2009, 15:46
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Hades wrote:
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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Re: Tough DS 6 [#permalink]

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New post 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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Re: Tough DS 6 [#permalink]

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New post 03 Jun 2009, 16:16
Amazing problem and explanation Hades.
+1
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Re: Tough DS 6   [#permalink] 03 Jun 2009, 16:16
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