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DS: Kaplan CAT Question

by jaroth04 » Sat Dec 05, 2009 1:29 pm
Just finished a Kaplan prep test, that contained the following question:

Are p, q, and r all even integers?

(1) p + 2 = q = r - 2
(2) pqr = 0

See spoiler text for OA, as well as my question.

[spoiler]
OA is C (BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient). I chose B (Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient).

What I inferred (correctly from the official explanation, I believe) is that (2) implies that either p, q or r must be 0. 0 is not, to my understanding, an even integer. Can someone refute that? Is 0 considered to be an even integer on the GMAT? Or am I misreading or misinterpreting the statement? If one of them is 0, then they must not *all* be even integers, which would mean that the answer should be B?
[/spoiler]

Thanks in advance for your input.

-Joe
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by KapTeacherEli » Sat Dec 05, 2009 5:36 pm
0 is even. An even number is an integer that, when divided by two, results in an integer. 0/2 = 0.

If it's any consolation, I made this mistake on the GMAT when i was testing to be a Kaplan teacher; it's not immediately obvious that 0 should be even. But it is!

Best of luck with your studies.
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www.kaptest.com/gmat

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by Stuart@KaplanGMAT » Sun Dec 06, 2009 10:50 am
Every integer is either even or odd.
Every integer is either negative, 0 or positive.

So, 0 is an even, "uncharged", integer.

Here's a good way to reason out why 0 must be even: out of every pair of consecutive integers, one is even and one is odd. If 0 weren't even, then when we get to {-1, 0, 1} we'd violate that rule.

Another way to remember is that oddness and evenness are determined by the last digit in an integer; since 10, 20, 30, 1000 and so on are all even, so must be 0.
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by Gmatter2.0 » Sun Dec 06, 2009 6:26 pm
Is my Procedure Correct.

Statement 1:
p+2=q=r-2

This gives no additional information p can be a fraction or odd and still q remains valid and so does r.
A and D are eliminated
Statement 2:
Give us interesting information, pqr product is zero,
It means either or all of p,q and r are zero.

But 2 by itself does not give any information, because other than Zero elements can be Odd or even .

Now combining 1 and 2

let p=0
q=2
and r=4

Let r=0
q=-2
p=-4

Let q=0
p=-2
r=0

Hence C is the correct choice....
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by Stuart@KaplanGMAT » Sun Dec 06, 2009 7:33 pm
Gmatter2.0 wrote:Is my Procedure Correct.

Statement 1:
p+2=q=r-2

This gives no additional information p can be a fraction or odd and still q remains valid and so does r.
A and D are eliminated
Statement 2:
Give us interesting information, pqr product is zero,
It means either or all of p,q and r are zero.

But 2 by itself does not give any information, because other than Zero elements can be Odd or even .

Now combining 1 and 2

let p=0
q=2
and r=4

Let r=0
q=-2
p=-4

Let q=0
p=-2
r=0

Hence C is the correct choice....
Your reasoning is correct, but we could have done a bit less work.

From (1), we can see that that p, q and r have a distance of 2 between each pair of numbers. Therefore, either they're all integers or none of them are integers. We don't know which one, so insufficient.

From (2), we know that one of them is 0, but we don't know anything about the other two, so insufficient.

Combined: from (2) we know that one of them has to be 0, an integer; add that info to (1) and we now know that all 3 of them have to be integers. Together sufficient, choose C.
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