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# If n is a positive integer is n-1 divisible by 3 ?

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If n is a positive integer is n-1 divisible by 3 ? [#permalink]

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24 Sep 2012, 23:09
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If n is a positive integer is n-1 divisible by 3 ?

(1) n^2+n is not divisible by 6.

(2) 3n=3k+3 where k is a positive multiple of 3.
[Reveal] Spoiler: OA
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Re: If n is a positive integer is n-1 divisible by 3 ? [#permalink]

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25 Sep 2012, 00:10
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harikris wrote:
Hi Guys,

Could you please a solution to this problem ?

If n is a positive integer is n-1 divisible by 3 ?

1) n^2+n is not divisible by 6.

2)3n=3k+3 where k is a positive multiple of 3.

Thanks,

harikris

If you look at any three consecutive integers, one of them will always be a multiple of 3, since multiples of 3 are exactly 3 apart.

From Statement 1, we know that (n)(n+1) is not divisible by 6. One of n or n+1 is even, since n and n+1 are consecutive integers, so (n)(n+1) must be divisible by 2. So if (n)(n+1) is not divisible by 6, it must not be divisible by 3, so neither n nor n+1 are divisible by 3. But n-1, n, and n+1 are three consecutive integers, and one of them must be divisible by 3. If n and n+1 are not, then n-1 must be, so Statement 1 is sufficient.

From Statement 2, if 3n = 3k + 3, then n = k + 1. So n-1 = k, and since k is a multiple of 3, so must be n-1, since they're the same number. So Statement 2 is also sufficient and the answer is D.

Either you've miscopied Statement 2 or the OA is not right.
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If n is a positive integer is n-1 divisible by 3 ? [#permalink]

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25 Sep 2012, 01:35
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If n is a positive integer is n-1 divisible by 3 ?

(1) n^2+n is not divisible by 6 --> $$n^2+n=n(n+1)$$ is not divisible by 3 (since it's not divisible by 6) --> neither $$n$$ nor $$n+1$$ is divisible by 3. Out of three consecutive integers $$n-1$$, $$n$$ and $$n+1$$, one must be divisible by 3, since we know that it's not $$n$$ or $$n+1$$, then it must be $$n-1$$. Sufficient.

(2) 3n=3k+3 where k is a positive multiple of 3 --> reduce by 3: $$n=k+1$$ --> $$n-1=k=multiple \ of \ 3$$. Sufficient.

Answer: D. (OA is not correct).
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Re: If n is a positive integer is n-1 divisible by 3 ? [#permalink]

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31 Oct 2012, 09:18
Bunuel wrote:
If n is a positive integer is n-1 divisible by 3 ?

(1) n^2+n is not divisible by 6 --> $$n^2+n=n(n+1)$$ is not divisible by 3 (since it's not divisible by 6) --> neither $$n$$ nor $$n+1$$ is divisible by 3. Out of three consecutive integers $$n-1$$, $$n$$ and $$n+1$$, one must be divisible by 3, since we know that it's not $$n$$ or $$n+1$$, then it must be $$n-1$$. Sufficient.

(2) 3n=3k+3 where k is a positive multiple of 3 --> reduce by 3: $$n=k+1$$ --> $$n-1=k=multiple \ of \3$$. Sufficient.

Answer: D. (OA is not correct).

The original question for statement 2) is : 3n=k+3 where k is a positive multiple of 3
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Re: If n is a positive integer is n-1 divisible by 3 ? [#permalink]

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01 Nov 2012, 07:18
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superpus07 wrote:
Bunuel wrote:
If n is a positive integer is n-1 divisible by 3 ?

(1) n^2+n is not divisible by 6 --> $$n^2+n=n(n+1)$$ is not divisible by 3 (since it's not divisible by 6) --> neither $$n$$ nor $$n+1$$ is divisible by 3. Out of three consecutive integers $$n-1$$, $$n$$ and $$n+1$$, one must be divisible by 3, since we know that it's not $$n$$ or $$n+1$$, then it must be $$n-1$$. Sufficient.

(2) 3n=3k+3 where k is a positive multiple of 3 --> reduce by 3: $$n=k+1$$ --> $$n-1=k=multiple \ of \3$$. Sufficient.

Answer: D. (OA is not correct).

The original question for statement 2) is : 3n=k+3 where k is a positive multiple of 3

In this case for (2) we'll have: 3n=k+3 where k is a positive multiple of 3 --> if k=3, then n=2 and n-1 is not divisible by 3 but if k=9 then n=4 and n-1 is divisible by 3. Not sufficient.
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Re: If n is a positive integer is n-1 divisible by 3 ? [#permalink]

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06 Nov 2014, 01:29
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Re: If n is a positive integer is n-1 divisible by 3 ?   [#permalink] 06 Nov 2014, 01:29
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