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# OG 11

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Joined: 05 Mar 2011
Posts: 155
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Kudos [?]: 16 [0], given: 3

OG 11 [#permalink]  10 Dec 2011, 16:46
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Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 2 sessions
147. IF N IS A POSITIVE INTEGER, IS N^3-N DIVISIBLE BY 4.

1. N = 2K + 1, WHERE K IS AN INTEGER.

WHY IS THIS SUFFICIENT -I GOT THIS WRONG :/
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Joined: 24 Jun 2008
Posts: 978
Location: Toronto
Followers: 281

Kudos [?]: 799 [0], given: 3

Re: OG 11 [#permalink]  10 Dec 2011, 18:19
ashiima wrote:
147. IF N IS A POSITIVE INTEGER, IS N^3-N DIVISIBLE BY 4.

1. N = 2K + 1, WHERE K IS AN INTEGER.

WHY IS THIS SUFFICIENT -I GOT THIS WRONG :/

We can factor the expression in the question, using the difference of squares:

n^3 - n = n(n^2 - 1) = (n)(n-1)(n+1)

Arranging these factors in order, we want to know if (n-1)(n)(n+1) is divisible by 4. Now these three factors are consecutive integers since they are 1 apart. If n is odd, as Statement 1 tells us, then the other two factors, n-1 and n+1, must both be even, so they each must be divisible by 2, and their product must be divisible by 2*2 = 4.

While we don't need it for this question, it's actually true that (n-1)(n)(n+1) must be divisible by 8 if n is odd, since then n-1 and n+1 are consecutive even numbers, and whenever you have two consecutive even numbers, one of them must be a multiple of 4, and the other a multiple of 2.
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Re: OG 11   [#permalink] 10 Dec 2011, 18:19
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