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

Question

If  Y  is a factor of  Z , is  X  even?

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

(2)  XY  is even

Neochronic · Answer

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..

mjGMAT · Answer

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.

pmenon · Answer

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

Hades · Answer

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 .

lahoosaher · Answer

Hello mjGMAT,

if

z/y=k I think you have used z/y = yk

lahoosaher · Answer

Amazing problem and explanation Hades.
+1

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

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