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# M10-09

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Math Expert
Joined: 02 Sep 2009
Posts: 51185

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15 Sep 2014, 23:41
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Difficulty:

25% (medium)

Question Stats:

68% (00:53) correct 32% (00:32) wrong based on 88 sessions

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If $$n$$ is a positive integer is $$n(n+1)$$ divisible by 6?

(1) $$n$$ is a multiple of 2

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

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Joined: 02 Sep 2009
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15 Sep 2014, 23:41
Official Solution:

(1) $$n$$ is a multiple of 2. If $$n=2$$ then the answer is YES but if $$n=4$$ then the answer is NO. Not sufficient.

(2) $$n$$ is a multiple of 3. Notice that $$n(n+1)$$ is a product of two consecutive integers hence one of them must be even so $$n(n+1)$$ is divisible by 2 and since $$n$$ is a multiple of 3 then $$n(n+1)$$ is divisible 3 too, so $$n(n+1)$$ is divisible by $$2*3=6$$. Sufficient.

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22 Nov 2017, 10:18
Hi Bunuel,

I have a question here, we are given that n is positive integer i.e. n>= 1.

If n = 1 then n(n+1) will be 1*2 , which is not divisible by 6. In this case this is not divisible by 3 but only with 2.

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22 Nov 2017, 10:21
msk0657 wrote:
Hi Bunuel,

I have a question here, we are given that n is positive integer i.e. n>= 1.

If n = 1 then n(n+1) will be 1*2 , which is not divisible by 6. In this case this is not divisible by 3 but only with 2.

Which statement are you talking about???

If (2), then we are given that n is a multiple of 3. Is 1 a multiple of 3???
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22 Nov 2017, 10:27
msk0657 wrote:
Hi Bunuel,

I have a question here, we are given that n is positive integer i.e. n>= 1.

If n = 1 then n(n+1) will be 1*2 , which is not divisible by 6. In this case this is not divisible by 3 but only with 2.

Hi msk0657

not all two consecutive integers are divisible by 3 and hence by 6 eg. (4,5) (7,8) etc.
here n is a multiple of 3, hence n(n+1) is divisible by 6 because both factors of 6 = 2 & 3 are present in the product n(n+1)
Re: M10-09 &nbs [#permalink] 22 Nov 2017, 10:27
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# M10-09

Moderators: chetan2u, Bunuel

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