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They say that n(n-2) becomes divisible by 4 as soon as n is even, so n must be odd. Well what about n=2? -->Isn't n^2-2n=0, an even integer? Did Manhattan Gmat forget about that option?

They say that n(n-2) becomes divisible by 4 as soon as n is even, so n must be odd. But I believe n could be 2 such that the result is even --> 0.

Oh!!! n^2-2n will be divisible by 4 for all even n's. But statement 1 says that the expression is not divisible by 4. Thus, "n" is definitely not even; all integers are either even or odd; if n is not even, it is odd.

Thus, the answer to the question is: Yes, "n" is odd. And statement 1 is sufficient. ****************************

If n=2; n(n-2)=2*0=0; "0" is divisible by 4. Thus, n=2 doesn't fit well with the condition given in statement 1.

Note: 0 is even. 0 is divisible by all real numbers but 0 itself. 0 is a multiple of all real numbers.
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(1) SUFFICIENT:\(n^2\) – 2n = n(n – 2). If n is even, both terms in this product will be even, and the product will be divisible by 4. Since n2 – 2n is not a multiple of 4, we know that the integer n cannot be even—it must be odd.

(2) INSUFFICIENT: Multiples of 3 can be either odd or even.

Here's my question: With regards to (1), n could be 2, in which case 2(2-2)=0, in which case the expression is not a multiple of 4, but n is even. Thus shouldn't the answer be E?

Re: Is the integer n odd? (1) n^2 – 2n is not a multiple of 4. [#permalink]

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13 Nov 2013, 10:51

AccipiterQ wrote:

Here's my question: With regards to (1), n could be 2, in which case 2(2-2)=0, in which case the expression is not a multiple of 4, but n is even. Thus shouldn't the answer be E?

0 is a multiple of every integer except zero itself.
_________________

(1) SUFFICIENT:\(n^2\) – 2n = n(n – 2). If n is even, both terms in this product will be even, and the product will be divisible by 4. Since n2 – 2n is not a multiple of 4, we know that the integer n cannot be even—it must be odd.

(2) INSUFFICIENT: Multiples of 3 can be either odd or even.

Here's my question: With regards to (1), n could be 2, in which case 2(2-2)=0, in which case the expression is not a multiple of 4, but n is even. Thus shouldn't the answer be E?

Re: Is the integer n odd? (1) n^2 – 2n is not a multiple of 4. [#permalink]

Show Tags

07 Jan 2014, 01:38

2

This post received KUDOS

1. \(n^2-2n\) is not a multiple of 4.

follows \(n(n-2)\)is not a multiple of 4

find out in which points the equation yields zero, zero is a multiple of 4 as well as a multiple of any number except zero itself. value a=0 and value b=2 thus n must not be any of those two values because otherwise the expression would result in a multiple of 4 -n can be both positive or negative, the only restriction is that n is an integer- since o and 2 are out of the pool 4 would be the first positive even number applicable to n. Every other positive even number would cause the expression to be a multiple of 4. We can safely say that n is for sure not an even number. before submitting the answer let's quickly check how the expression behaves with negatives.

if n=-2 the greatest negative even integer plugged \(n^2-2n\) results in a multiple of 4 then for sure n is odd.

Sufficient.

2. n is a multiple of 3. Non sufficient, n could be zero.
_________________

learn the rules of the game, then play better than anyone else.

(1) n²-2n ==> n(n-2) = 0. Thus, n = 2 or n = 0. Both are multiples of 4. Every other even integer (also the negatives) result in a multiple of 4. Thus n is clearly odd. You could also just plug in numbers....

(2) n is a multiple of 3 --> clearly IS. could be 6 or 9 or 12 or 15 and so on.

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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(1) n^2-2n is not a multiple of 4 (2) n is a multiple of 3

Given : n is an Integer

Question : Is n odd?

Statement 1: n^2-2n is not a multiple of 4 i.e. n(n-2) is a multiple of 4 but since n and (n-2) are separated by 2 therefore, they will both be either even or both be odd Since the product is even so they must be even. Hence SUFFICIENT

Statement 2: n is a multiple of 3 a multiple of 3 may be even (e.g. 6 or 12) or may be odd (e.g. 3 or 9). Hence, NOT SUFFICIENT

(1) n^2-2n is not a multiple of 4 (2) n is a multiple of 3

Given : n is an Integer

Question : Is n odd?

Statement 1: n^2-2n is not a multiple of 4 i.e. n(n-2) is a multiple of 4 but since n and (n-2) are separated by 2 therefore, they will both be either even or both be odd Since the product is even so they must be even. Hence SUFFICIENT

Statement 2: n is a multiple of 3 a multiple of 3 may be even (e.g. 6 or 12) or may be odd (e.g. 3 or 9). Hence, NOT SUFFICIENT

Answer: Option A

Very good explanation....Thx +1 Kudos

gmatclubot

Re: Is the integer n odd?
[#permalink]
26 Jan 2016, 06:40

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