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## Is zero considered an even integer?

tagged by: Brent@GMATPrepNow

This topic has 5 expert replies and 6 member replies
givemeanid Master | Next Rank: 500 Posts
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#### Is zero considered an even integer?

Sun Jul 01, 2007 5:59 pm
GMAT considers 0 as an even integer. Is that correct?

800_or_bust Master | Next Rank: 500 Posts
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Wed May 04, 2016 6:49 pm
Matt@VeritasPrep wrote:
I'd go even further and say that if you see the word 'nonnegative' instead of 'positive', considering 0 is *almost definitely* crucial to solving the problem.

(This goes for 'nonpositive' too, I suppose, though I can't remember seeing that term in a GMAT problem.)
Probably should remind the other instructors of this. I was watching the recorded stream from last night's free sample class, and noticed an error in the first example.

The prompt was "If n is a nonnegative integer, then n(n+1)(n+2) is ..." And the correct answer was given as (E) Divisible by 12 whenever n is even.

But this is clearly not correct, because if n=0, then n(n+1)(n+2) evaluates to be 2 which is not divisible by 12. Zero is both a nonnegative integer (as required by the prompt) and even (as required in the selected answer choice), and proves that (E) is not necessarily true.

Other than that, it was good. Brian was pretty entertaining. Seemed like he was battling a bit of a cough though!

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800_or_bust Master | Next Rank: 500 Posts
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Thu May 05, 2016 3:05 am
800_or_bust wrote:
Matt@VeritasPrep wrote:
I'd go even further and say that if you see the word 'nonnegative' instead of 'positive', considering 0 is *almost definitely* crucial to solving the problem.

(This goes for 'nonpositive' too, I suppose, though I can't remember seeing that term in a GMAT problem.)
Probably should remind the other instructors of this. I was watching the recorded stream from last night's free sample class, and noticed an error in the first example.

The prompt was "If n is a nonnegative integer, then n(n+1)(n+2) is ..." And the correct answer was given as (E) Divisible by 12 whenever n is even.

But this is clearly not correct, because if n=0, then n(n+1)(n+2) evaluates to be 2 which is not divisible by 12. Zero is both a nonnegative integer (as required by the prompt) and even (as required in the selected answer choice), and proves that (E) is not necessarily true.

Other than that, it was good. Brian was pretty entertaining. Seemed like he was battling a bit of a cough though!
Actually, I'm the moron. I guess it was too late to be doing anything GMAT related last night... Of course, that expression actually evaluates to zero, not 2. And yes, zero is divisible by 12 so (E) is always true. Zero is divisible by any number. Oops.

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jayhawk2001 Community Manager
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Sun Jul 01, 2007 7:50 pm
Well, not just GMAT . Zero is generally considered even.

An integer n is called even if there exists an integer m such that n = 2m, and odd if n+1 is even.

montz Senior | Next Rank: 100 Posts
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Mon Aug 20, 2007 10:34 am
I always thought zero is neither odd nor even. The properties of even/odd nos. -

Even => 2n and Odd => 2n-1 where 'n' is a natural number.

But as per OG zero is even

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Fri Aug 31, 2007 4:53 am
however one thing to note is that zero is neither +ve nor -ve

samirpandeyit62 Master | Next Rank: 500 Posts
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Wed Sep 05, 2007 9:49 am
As far as sign of 0 is concerned it is generally accepted that 0 can be both +ve as well as -ve as +0 =0 & -0 = 0 hence both would be the same nos,
I believe specially on a DS question if it is asked to check some value is +ve or -ve and if it comes out to be 0 or can possibly be 0, I think we can simply use it as required by the question & not against it i.e if question wants +ve value, use it as +ve and vice versa.

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Brent@GMATPrepNow GMAT Instructor
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Thu Apr 07, 2016 6:00 am
alstonamos wrote:
I always thought zero is neither odd nor even. The properties of even/odd nos. -

waleeed
Zero is definitely even. In fact, this concept is tested often on the GMAT.
An integer is considered even if it can be written as the product of 2 and another integer.
Since, 0 = (2)(0), zero is considered even.

Zero, however, is neither positive nor negative.

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Brent

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Thu Apr 07, 2016 5:26 pm
Hi Waleed,

0 is called a 'null value' since it's neither positive nor negative.
0 is even because on a number line, integers vary odd-even-odd-even-etc. (e.g.. -3, -2, -1, 0, 1, 2, etc.)

When a question uses a phrase such as "X is a non-negative integer', you should ABSOLUTELY consider the possibility that X is 0.

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Fri Apr 15, 2016 1:38 pm
I'd go even further and say that if you see the word 'nonnegative' instead of 'positive', considering 0 is *almost definitely* crucial to solving the problem.

(This goes for 'nonpositive' too, I suppose, though I can't remember seeing that term in a GMAT problem.)

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Thu May 26, 2016 1:47 pm
800_or_bust wrote:
The prompt was "If n is a nonnegative integer, then n(n+1)(n+2) is ..." And the correct answer was given as (E) Divisible by 12 whenever n is even.

But this is clearly not correct, because if n=0, then n(n+1)(n+2) evaluates to be 2 which is not divisible by 12. Zero is both a nonnegative integer (as required by the prompt) and even (as required in the selected answer choice), and proves that (E) is not necessarily true.

Other than that, it was good. Brian was pretty entertaining. Seemed like he was battling a bit of a cough though!
Sorry to have missed this response!

If n is even, you'll have n even, (n + 2) even, and one of the three consecutive integers divisible by 3 (though you don't know which one it will be). So you have two factors of 2 and one factor of 3, making n * (n + 1) * (n + 2) always divisible by 12.

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Matt@VeritasPrep GMAT Instructor
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Thu May 26, 2016 1:49 pm
We could also think of this algebraically. Suppose n is even, so we can write n = 2k, where k is some integer we don't care about.

2k * (2k + 1) * (2k + 2) =>

4k * (2k + 1) * (k + 1)

So clearly this is divisible by 4. We're left with

k * (k + 1) * (2k + 1)

If k has remainder 0 by 3, then the k term divides by 3.

If k has remainder 1 by 3, then the (2k + 1) term divides by 3.

If k has remainder 2 by 3, then the (k + 1) term divides by 3.

Since these three cases are exhaustive, our product must divide by 3 and by 4, and hence by 12.

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