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# Is positive integer n 1 a multiple of 3?

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Is positive integer n 1 a multiple of 3? [#permalink]  25 Sep 2010, 10:06
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Question Stats:

51% (02:01) correct 49% (01:35) wrong based on 49 sessions
Is positive integer n – 1 a multiple of 3?

(1) n^3 – n is a multiple of 3

(2) n^3 + 2n^2 + n is a multiple of 3

[Reveal] Spoiler:
For 1, for following values of n, $$n^3 - n$$ becomes multiple of 3 : 2,3,4,5,6,7,8....

However, by plugging these values in n-1, we have different results. thus, not sufficient

For 2, following values of n gets the equation as a multiple of 3 :
2,3,5,6,7,8....
Its again not sufficient.

Combining 1 & 2, we again can't be conclusive. Thus,answer should be E.

Please write where am I wrong?
[Reveal] Spoiler: OA
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Kudos [?]: 574 [1] , given: 25

Re: Positive integer n-1 multiple of 3? [#permalink]  25 Sep 2010, 10:15
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KUDOS
Orange08 wrote:
Is positive integer n – 1 a multiple of 3?

(1) $$n^3 – n$$ is a multiple of 3

(2) $$n^3 + 2n^2+ n$$ is a multiple of 3

For 1, for following values of n, $$n^3 - n$$ becomes multiple of 3 : 2,3,4,5,6,7,8....

However, by plugging these values in n-1, we have different results. thus, not sufficient

For 2, following values of n gets the equation as a multiple of 3 :
2,3,5,6,7,8....
Its again not sufficient.

Combining 1 & 2, we again can't be conclusive. Thus,answer should be E.

Please write where am I wrong?

(1) $$n^3-n = n(n^2-1) = n(n+1)(n-1)$$
This number will always be a multiple of three ... so not sufficient to answer our question

(2) $$n^3 + 2n^2+ n = n(n+1)^2$$
If this is a multiple of 3, then either n is a multiple of 3 or (n+1) is a multiple of 3. In either case, n-1 cannot be a multiple of 3.
Hence this is sufficient

In your solution above, notice that for all the n's in case (2) n-1 is not a multiple of 3.
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Re: Positive integer n-1 multiple of 3? [#permalink]  25 Sep 2010, 10:16
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Expert's post
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BOOKMARKED
Orange08 wrote:
Is positive integer n – 1 a multiple of 3?

(1) n^3 – n is a multiple of 3

(2) $$n^3 + 2n^2+ n$$ is a multiple of 3

For 1, for following values of n, $$n^3 - n$$ becomes multiple of 3 : 2,3,4,5,6,7,8....

However, by plugging these values in n-1, we have different results. thus, not sufficient

For 2, following values of n gets the equation as a multiple of 3 :
2,3,5,6,7,8....
Its again not sufficient.

Combining 1 & 2, we again can't be conclusive. Thus,answer should be E.

Please write where am I wrong?

Is positive integer n – 1 a multiple of 3?

(1) n^3 – n is a multiple of 3 --> $$n^3-n=n(n^2-1)=(n-1)n(n+1)=3q$$. Now, $$n-1$$, $$n$$, and $$n+1$$ are 3 consecutive integers and one of them must be multiple of 3, so no wonder that their product is a multiple of 3. However we don't know which one is a multiple of 3. Not sufficient.

(2) $$n^3 + 2n^2+ n$$ is a multiple of 3 --> $$n^3 + 2n^2+ n=n(n^2+2n+1)=n(n+1)^2=3p$$ --> so either $$n$$ or $$n+1$$ is a multiple of 3, as out of 3 consecutive integers $$n-1$$, $$n$$, and $$n+1$$ only one is a multiple of 3 then knowing that it's either $$n$$ or $$n+1$$ tells us that $$n-1$$ IS NOT multiple of 3. Sufficient.

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Re: Positive integer n-1 multiple of 3?   [#permalink] 25 Sep 2010, 10:16
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