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# Is the integer n odd?

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Is the integer n odd?  [#permalink]

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Updated on: 02 Jun 2011, 04:16
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Question Stats:

63% (01:25) correct 37% (01:46) wrong based on 258 sessions

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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

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?

Originally posted by heyholetsgo on 01 Jun 2011, 15:43.
Last edited by heyholetsgo on 02 Jun 2011, 04:16, edited 2 times in total.
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01 Jun 2011, 15:52
heyholetsgo wrote:
Is the integer n odd?

1.) n^2-2n
2.) n is a multiple of 3
Well what about n=2? --> n^2-2n=0, an even integer? Did Manhattan Gmat forget about that option?

Guess you missed the RHS in statement 1:

1.) n^2-2n=0

$$n^2-2n=0$$
$$n(n-2)=0$$

Thus,
Either n=2 or n=0. Both are even.

We can conclusively say that "No, n is not odd"
Sufficient.

2)
n can be 3
OR
n can be 6.
Not Sufficient.

Ans: "A"
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01 Jun 2011, 16:52
1. Sufficient

n^2-2n =0

=> n=0 or n =2

in both the situations n is not odd and enough to answer the question

2. Not sufficient

as n can be 0 or 3 or 6 or 9....

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01 Jun 2011, 19:13
Yes, Clear A. B can be odd and even.
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02 Jun 2011, 04:15
Uhhh, I'm sorry guys, forgot a tiny but important part of the question....
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.
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02 Jun 2011, 04:37
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1
heyholetsgo wrote:
Uhhh, I'm sorry guys, forgot a tiny but important part of the question....
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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02 Jun 2011, 06:18
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Damn, 0 is divisible by 4. Thanks man;)
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19 Oct 2011, 01:35
fluke wrote:
Note:
0 is even.
0 is divisible by all real numbers but 0 itself.
0 is a multiple of all real numbers.

Is this correct? I would have thought a number is even only if it is divisible by 2 without there being any remainder.

*Edit*
I retract my question. Since even+even=even, zero must be an even number. Sorry for the confusion!
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20 Oct 2011, 23:55
hey

but how can u equate statement 1 to zero ? Its not given rite ?

Am I missing some thing silly ?
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21 Oct 2011, 00:48
Priyanka2011 wrote:
hey

but how can u equate statement 1 to zero ? Its not given rite ?

Am I missing some thing silly ?

We can't equate it to 0. My first reply was to an incomplete question. My second reply was the valid one.

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

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13 Nov 2013, 11:45
Is the integer n odd?

(1) $$n^2$$ – 2n is not a multiple of 4.
(2) n is a multiple of 3.

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

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07 Jan 2014, 02:38
2
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?  [#permalink]

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24 Jan 2014, 00:39
1
(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?  [#permalink]

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07 Jul 2015, 14:34
1
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

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Re: Is the integer n odd?  [#permalink]

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26 Jan 2016, 07:40
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

Very good explanation....Thx
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Re: Is the integer n odd?  [#permalink]

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22 Feb 2017, 12:38
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]

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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

Cheers,
Brent
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Re: Is the integer n odd?  [#permalink]

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04 Mar 2019, 10:24
#1
n^2-2n is not a multiple of 4
n(n-2)/4 not a multiple of 4
this will be true for n Odd integer values
hence sufficient
#2
n is a multiple of 3 ; n = 3,6,9,12
not sufficient
IMO A

heyholetsgo wrote:
Is the integer n odd?

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

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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Is the integer n odd?  [#permalink]

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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

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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Is the integer n odd?   [#permalink] 15 Mar 2019, 07:45
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