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# Is the integer n odd?

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Manager
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Is the integer n odd? [#permalink]  01 Jun 2011, 14:43
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Question Stats:

58% (02:10) correct 42% (01:18) wrong based on 128 sessions
Is the integer n odd?

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

[Reveal] Spoiler:
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?
[Reveal] Spoiler: OA

Last edited by heyholetsgo on 02 Jun 2011, 03:16, edited 2 times in total.
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heyholetsgo wrote:
Is the integer n odd?

1.) n^2-2n
2.) n is a multiple of 3
[Reveal] Spoiler:
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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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....

Manager
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Yes, Clear A. B can be odd and even.
Manager
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Uhhh, I'm sorry guys, forgot a tiny but important part of the question....
[Reveal] Spoiler:
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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heyholetsgo wrote:
Uhhh, I'm sorry guys, forgot a tiny but important part of the question....
[Reveal] Spoiler:
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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Damn, 0 is divisible by 4. Thanks man;)
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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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hey

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

Am I missing some thing silly ?
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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]  13 Nov 2013, 10:45
Is the integer n odd?

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

[Reveal] 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]  13 Nov 2013, 10:51
Expert's post
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]  13 Nov 2013, 10:53
Expert's post
AccipiterQ wrote:
Is the integer n odd?

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

[Reveal] 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?

Merging similar topics.
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Re: Is the integer n odd? (1) n^2 – 2n is not a multiple of 4. [#permalink]  07 Jan 2014, 01:38
1
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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? [#permalink]  23 Jan 2014, 23:39
(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]  07 Jul 2015, 02:13
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Re: Is the integer n odd? [#permalink]  07 Jul 2015, 13:34
Expert's post
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] 07 Jul 2015, 13:34
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