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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, 11:51

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, 02:38

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.
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(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

Answer: Option A
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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

We don't need to know which value n might be, just whether n is odd. Therefore, do not rephrase this question to “What is integer n?” Doing so unnecessarily increases the amount of information we need to answer the question. Of course, if you happen to know what n is, then great, you can answer any Yes/No question about n. But you generally don't need to know the value of n to answer Yes/No questions about n. (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 n^2 – 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.

Hence A.
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Re: Is the integer n odd?
[#permalink]
22 Feb 2017, 12:38

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