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Is the integer n odd? (1) n^2 - 2n is not a multiple of 4 (2) n is a

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heyholetsgo
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Is the integer n odd? (1) n^2 - 2n is not a multiple of 4 (2) n is a [#permalink] New post   Updated on: 28 Mar 2021, 11:06
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Is the integer n odd?

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



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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?
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A

Originally posted by heyholetsgo on 01 Jun 2011, 15:43.
Last edited by Bunuel on 28 Mar 2021, 11:06, edited 4 times in total.
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Re: Is the integer n odd? (1) n^2 - 2n is not a multiple of 4 (2) n is a [#permalink] New post   02 Jun 2011, 06:18
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Damn, 0 is divisible by 4. Thanks man;)
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Re: Is the integer n odd? (1) n^2 - 2n is not a multiple of 4 (2) n is a [#permalink] New post   07 Jan 2014, 02:38
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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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Re: Is the integer n odd? (1) n^2 - 2n is not a multiple of 4 (2) n is a [#permalink] New post   24 Jan 2014, 00:39
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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.
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Re: Is the integer n odd? (1) n^2 - 2n is not a multiple of 4 (2) n is a [#permalink] New post   07 Jul 2015, 14:34
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heyholetsgo wrote:
Is the integer n odd?

(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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Re: Is the integer n odd? (1) n^2 - 2n is not a multiple of 4 (2) n is a [#permalink] New post   20 Apr 2018, 07:52
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heyholetsgo wrote:
Is the integer n odd?

(1) n² - 2n is not a multiple of 4
(2) n is a multiple of 3


Target question: Is the integer n odd?

Given: n is an INTEGER

Statement 1: n² - 2n is not a multiple of 4.
Factor to get: n(n - 2) is NOT a multiple of 4

Underlying concepts:

Integer n is 2 greater than n-2
If n is ODD, then n-2 is also ODD, so n(n - 2) = (ODD)(ODD) = ODD. In this case, n(n-2) cannot be divisible by 4
If n is EVEN, then n-2 is also EVEN, so n(n - 2) = (EVEN)(EVEN) = EVEN. More importantly, n and n-2 are CONSECUTIVE even integers, and the product of two CONSECUTIVE even integers is always a multiple of 4

Statement 1 tells us that n(n - 2) is NOT a multiple of 4
So, it cannot be the case that n is EVEN
In other words, it MUST be the case that n is ODD
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: n is a multiple of 3
There are several values of n that satisfy this condition. Here are two:
Case a: n = 3, in which case n n is ODD
Case b: n = 6, in which case n n is EVEN
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

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Re: Is the integer n odd? (1) n^2 - 2n is not a multiple of 4 (2) n is a [#permalink] New post   15 Mar 2019, 07:45
GMATinsight wrote:
heyholetsgo wrote:
Is the integer n odd?

(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


There's a slight mistake (bolded in red) there. If both n & (n-2) are even (example: {2,0} or {4,2}), then they would always be a multiple of 4, violating stmt 1. Hence, they both have to be odd. Therefore, n = odd.
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Re: Is the integer n odd? (1) n^2 - 2n is not a multiple of 4 (2) n is a [#permalink] New post   28 Mar 2021, 05:34
Bunuel, I think the question should get proper formatting.

I tried out different values and didn't get a sufficient answer from A, because I thought that the argument is \(n^{2-2n}\).
When n = 1, \(n^{2-2n}\) becomes \(1^0 = 1\), which isn't a multiple of 4 and n is odd.
But when you plug in n = 2, it becomes \(2^{2-2*2} = 2^{2-4} = 2^{-2} = \frac{1}{4}\), which isn't a multiple of 4 as well and n is even.
Therefore 2 different answers -> insufficient.

It would be much clearer if the question was written as \(n^2-2n\)
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Re: Is the integer n odd? (1) n^2 - 2n is not a multiple of 4 (2) n is a [#permalink] New post   28 Mar 2021, 11:06
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argerniki wrote:
Bunuel, I think the question should get proper formatting.

I tried out different values and didn't get a sufficient answer from A, because I thought that the argument is \(n^{2-2n}\).
When n = 1, \(n^{2-2n}\) becomes \(1^0 = 1\), which isn't a multiple of 4 and n is odd.
But when you plug in n = 2, it becomes \(2^{2-2*2} = 2^{2-4} = 2^{-2} = \frac{1}{4}\), which isn't a multiple of 4 as well and n is even.
Therefore 2 different answers -> insufficient.

It would be much clearer if the question was written as \(n^2-2n\)


n^2 - 2n can only mean \(n^2 - 2n\). If it were \(n^{2 - 2n}\) it would have been written as n^(2 - 2n). Still edited. Thank you.
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Re: Is the integer n odd? (1) n^2 - 2n is not a multiple of 4 (2) n is a [#permalink] New post   10 Aug 2022, 08:53
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Re: Is the integer n odd? (1) n^2 - 2n is not a multiple of 4 (2) n is a [#permalink]
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