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If n is a positive integer, is n^3 - n divisible by 4? (1) n

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If n is a positive integer, is n^3 - n divisible by 4? (1) n [#permalink]

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New post 18 Jun 2011, 22:09
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Question Stats:

64% (01:26) correct 36% (02:09) wrong based on 76 sessions

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If n is a positive integer, is n^3 - n divisible by 4?

(1) n = 2k+1, where k is an integer
(2) n^2 + n is divisible by 6

[Reveal] Spoiler:
I actually got this answer wrong, but wanted to ask the forum if my thinking is incorrect.

So here was my thought process.

n^3 - n = n(n^2 -n) = n(n+1)(n-1), so if n is odd and greater than 1, then the equation is divisible by 4.

For (1), n can be 3 if k=1, but n can be 1 if k=0, since 0 is an integer. So this is insufficient.
For (2), n^2 + n = n(n+1). n can be many numbers and still be divisible by 6, so this is insufficient.

I chose E, but the answer is A. The OG states that for (1), 2k is even, so 2k+1 must be at least 3. My question is, isn't 0 an integer? If that is the case, then 2k does not necessarily have to be even, since it can be 0.

Thanks in advance for the help.


OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/if-n-is-a-po ... 09941.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 14 Mar 2017, 09:52, edited 2 times in total.
Edited.

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Re: If n is a positive integer, is n^3 - n divisible by 4? (1) n [#permalink]

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New post 18 Jun 2011, 23:00
btm3 wrote:
If n is a positive integer, is n^3 - n divisible by 4?
(1) n = 2k+1, where k is an integer
(2) n^2 + n is divisible by 6

I actually got this answer wrong, but wanted to ask the forum if my thinking is incorrect.

So here was my thought process.

n^3 - n = n(n^2 -n) = n(n+1)(n-1), so if n is odd and greater than 1, then the equation is divisible by 4.

For (1), n can be 3 if k=1, but n can be 1 if k=0, since 0 is an integer. So this is insufficient.
For (2), n^2 + n = n(n+1). n can be many numbers and still be divisible by 6, so this is insufficient.

I chose E, but the answer is A. The OG states that for (1), 2k is even, so 2k+1 must be at least 3. My question is, isn't 0 an integer? If that is the case, then 2k does not necessarily have to be even, since it can be 0.

Thanks in advance for the help.


The answer is indeed A.
St1: n = 2k+1, where k is an integer

2k has to be even , +1 makes its odd
so n = odd

question says n^3-n/4 is an integer
this is true for any odd value of n .. st1 gives us that .. hence sufficient.

your approach is right as well:
N^3-n = n(n^2-1) = n(n+1)(n-1) /4 is a integer
which means we need to have atleast 2 even numbers ( which can give atleast two 2's)
thus n must be odd, as n+1 = 4 and n-1 = 2
suppose n is even( ex :2) then n+1 = 3 and n-1 = 1
then n(n+1)(n-1) /4 is not an integer

St2: n^2 + n is divisible by 6
n(n+1) /6 is an integer
n can be 6 ........ n^3-n/4 is not an integer
n can be 8.......... n^3 -n/4 is an integer
hence in sufficient

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Re: If n is a positive integer, is n^3 - n divisible by 4? (1) n [#permalink]

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New post 19 Jun 2011, 01:58
Thanks for the reply.

For (1), I thought that if k=0, then n = 2k+1 = 2(0)+1 =1. So if n=1, then n-1=0 and n+1=2, and as a result, 0x1x2 would not be divisible by 4. That's why I thought (1) was insufficient. Maybe I'm overthinking this.

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Re: If n is a positive integer, is n^3 - n divisible by 4? (1) n [#permalink]

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Re: If n is a positive integer, is n^3 - n divisible by 4? (1) n [#permalink]

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New post 14 Mar 2017, 07:25
The first statement is sufficient. The expression n^3-n indicates that a product of 3 consecutive integers is being sought. The simplified version of the expression is n (n-1)(n+1).

The first statement indicates that n is an odd integer. Place any odd integer in the sequence and it is easy to derive that the expression is divisible by 4

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Re: If n is a positive integer, is n^3 - n divisible by 4? (1) n [#permalink]

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New post 14 Mar 2017, 09:52
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If n is a positive integer, is n^3 – n divisible by 4 ?

n^3-n=n(n^2-1)=(n-1)n(n+1), so we are asked whether the product of 3 consecutive integers is divisible bu 4.

(1) n = 2k + 1, where k is an integer --> n=odd --> as n is odd then both n-1 and n+1 are even hence (n-1)n(n+1) is divisible by 4. Sufficient.

(2) n^2 + n is divisible by 6 --> if n=2 then n^3-n=6 and the answer is NO but if n=3 then n^3-n=24 and the answer is YES. Not sufficient.

Answer: A.


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Re: If n is a positive integer, is n^3 - n divisible by 4? (1) n   [#permalink] 14 Mar 2017, 09:52
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