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OG Guide 12th Ed DS Q:170

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OG Guide 12th Ed DS Q:170 [#permalink] New post 24 Apr 2011, 07:09
If n is a positive integer, is n3 – n divisible by 4 ?
(1) n = 2k + 1, where k is an integer.
2) n2 + n is divisible by 6.

We recognize that this is 3 consecutive numbers
(n-1)(n)(n+1)

1) n=odd but if n=1 (0,1,2) the sum of the consecutive integers = 0, which is not div by 4 nor does (0,1,2) have two 2's.
I'm pulling my hair out trying to understand this.
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Re: OG Guide 12th Ed DS Q:170 [#permalink] New post 24 Apr 2011, 09:30
sterlinggrey wrote:
If n is a positive integer, is n3 – n divisible by 4 ?
(1) n = 2k + 1, where k is an integer.
2) n2 + n is divisible by 6.


Given: n is a positive integer.
To prove: (n^3-n)/4 = k (an integer)
(n-1)n(n+1)
Stmt1: n=2k+1. Substitute in above eqn: (2k+1-1)(2k+1)(2k+1+1)=(2k)(2k+1)(2k+2)=4k(2k+1)(k+1) which is divisible by 4. SUFF.
Stmt2: (n^2+n)/6 n(n+1)/6 take n=2 => (n-1)n(n+1)=1*2*3=6 not divisible by 4. take n=3 (n-1)n(n+1)=2*3*4=24 divisible by 4. Hence Insuff.

OA: A.
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Re: OG Guide 12th Ed DS Q:170 [#permalink] New post 24 Apr 2011, 12:44
DUUHHHH!

clearly I had reached my saturation point.
Thanks
Re: OG Guide 12th Ed DS Q:170   [#permalink] 24 Apr 2011, 12:44
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OG Guide 12th Ed DS Q:170

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