If n is positive integer, is (n^3 - n) divisible by 4? 1. n : DS Archive
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# If n is positive integer, is (n^3 - n) divisible by 4? 1. n

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

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08 Sep 2008, 13:34
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If n is positive integer, is (n^3 - n) divisible by 4?
1. n = 2K+1 where K is an interger
2. n^2 + n is divisible by 6

My query is that isn't Zero an integer?

If we plug in k = 0 in Stmt 1 we get n = 1 which is a +ve integer. Now if n=1, (n^3-n) = 0

is 0 divisible by 4?
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08 Sep 2008, 13:45
Yes, 0 is divisible by 4; think back to how you first learned division. If you have zero dollars, can you divide this amount up evenly among four people? Of course; everyone gets nothing. Indeed, 0 is divisible by every integer except for zero itself- you can never divide by zero.
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08 Sep 2008, 13:52
Ian,
That makes things a hell of a lot more clear now.

Thanks a ton.
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08 Sep 2008, 22:38
friends, the OA is what?
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09 Sep 2008, 00:39
I would go for A.

n^3-n = n(n^2-1) = (n-1) n (n+1) , the product of three consecutive integers. There are two ways for it to be divisible by 4:

- n is odd ==> (n-1), (n+1) are even ==> it's divisible by 2 twice ==> it's divisible by 4
- n is divisible by 4

A) is the former case, suff

B) n(n+1) is divisible by (3*2), not suff because n could be divisible by 2 and (n+1) be odd (es. if n=2, the number is not divisible by 4)
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09 Sep 2008, 00:39
stat 1) n= 1,3,5,7,9,11 etc . suff . given expression is divisble by 4

stat 2) n=2,3,5,6 etc
for n=3 yes
for n= 2 no. insuff.

A
Re: GmatPrep DS (n^3-n)   [#permalink] 09 Sep 2008, 00:39
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