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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] New post 18 Jun 2011, 21:09
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A
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C
D
E

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50% (01:54) correct 50% (02:01) wrong based on 4 sessions
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.
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Re: DS integer question [#permalink] New post 18 Jun 2011, 22:00
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: DS integer question [#permalink] New post 19 Jun 2011, 00:58
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.
Re: DS integer question   [#permalink] 19 Jun 2011, 00:58
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