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by binaras » Sat Mar 21, 2015 11:05 am
Hi

Need assistance in figuring out why the answer to the DS question (below) is option (D) Each statement alone is sufficient

"The positive integers x, y & z are such that x is a factor of y & y is a factor of z.
Is z even?

1) xz is even
2) y is even

Thanks

Thanks
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Source: — Data Sufficiency |

by Brent@GMATPrepNow » Sat Mar 21, 2015 11:35 am
The positive integers x, y, and z are such that x is a factor of y and y is a factor of z. Is z even?

(1) xz is even
(2) y is even.
Here's a more formal approach.

Target question: Is z even?

Given: x is a factor of y, and y is a factor of z.

There's a nice rule that says, "If D is a factor (divisor) of N, then N = kD for some integer k"
So, if x is a factor of y, then y = kx for some integer k.
Also, if y is a factor of z, then z = jy for some integer j

Statement 1: xz is even
This sets up two possible cases (x is even or z is even). We'll examine both:
case a: x is even.
If x is even, then kx is even, which means y is even (since y=kx).
If y is even, then jy is even, which means z is even (since z=jy).
case b: z is even
Since both possible cases yield the same answer to the target question, statement 1 is SUFFICIENT

Statement 2: y is even
If y is even, then jy is even, which means z is even (since z=jy).
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer = D

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by GMATGuruNY » Sat Mar 21, 2015 12:21 pm
The positive integers x, y and z are such that x is a factor of y and y is a factor of z. Is z even.

1) xz is even
2) y is even.
Let's reverse the language:

z is a multiple of y, and y is a multiple of x.
Thus, z is a multiple of x.

Statement 1: xz is even.
If x = even, then z must be even since it's a multiple of an even value.
If x = odd, then z must be even in order for xz to be even.
Since in either case z is even, sufficient.

Statement 2: y is even.
If z is a multiple of an even number, then z too must be even.
Sufficient.

The correct answer is D.
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by binaras » Sat Mar 21, 2015 7:37 pm
Thanks for the quick response. Appreciate it.

Binara
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