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# If n is +ve integer, is n*n*n - n divided by 4? 1) n=2k+1 ,

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Joined: 04 Sep 2008
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If n is +ve integer, is n*n*n - n divided by 4? 1) n=2k+1 , [#permalink]

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27 Oct 2008, 05:12
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If n is +ve integer, is n*n*n - n divided by 4?
1) n=2k+1 , where k is an integer
2) n*n + n is divided by 6

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rampuria

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Joined: 09 May 2008
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27 Oct 2008, 06:30
I would say A.

Consider this: n*n*n - n => n(n*n-1) => n(n-1)(n+1).
Look at this as the product of 3 consecutive integers: (n-1)n(n+1).
We're being asked if this product is divisible by 4, and it'll be if at least 2 terms are even (because this means that there are two 2's as a factor of this product).

1) n=2k+1
this tells us that n is odd (if k is even, say k=2, then n=5; if k is odd, say k=3, then n=7). Thus, (n-1) and (n+1) are even; if two terms are even, then n*n*n - n is divisible by 4 => SUFFICIENT

2) n*n + n is divisible by 6
this only means that the product n(n+1) has 2 & 3 as factors. However, we don't know which term is even and which one is odd, consequentely we don't what (n-1) is => INSUFFICIENT

Rampuria, where did you sourced did from?
OA?
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Joined: 05 Jul 2006
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27 Oct 2008, 09:14
If n is +ve integer, is n*n*n - n divided by 4?
1) n=2k+1 , where k is an integer
2) n*n + n is divided by 6

IS N(N^2 - 1) DEVISABLE BY 4

N(N+1)(N-1) 3 CONSECS.DEVISABLE BY 4 IF ( N-1) OR N+1 IS EVEN OR N IS DEVIABLE BY 4

FROM 1

N IS ODD...........SUFF

FROM 2

N(N+1) IS DEVIABLE BY 6

ONE OF THEM IS DEVIABLE BY 2 AS THE OTHER IS ODD........INSUFF

A

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This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

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Re: Number property DS   [#permalink] 27 Oct 2008, 09:14
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