DS Set 4 Q 11

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DS Set 4 Q 11

by arocks » Tue Oct 02, 2007 9:43 am
If n and k are positive integers, is n divisible by 6?
(1) n = k(k + 1)(k - 1)
(2) k -1 is a multiple of 3.

For statement 1, I understand that it is sufficient because -
n is a product of three consecutive terms .
And the sum of three consecutive terms is always divisible by 6
But if k=1, then n=0. In such a case the statement-1 is not sufficient.
Please let me know if i am missing something here. The question doesn’t mention that k>1.

Thanks.
Source: — Data Sufficiency |

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by kajcha » Tue Oct 02, 2007 10:07 am
IMO A

Stmt 1 is suff as product of 3 consecutive integers will have 6 as a factor.

Stmt 2 does not give relation between k and n so NOT SUFF

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by ratindasgupta » Wed Oct 03, 2007 6:22 am
Statement 1 is sufficient.

the rule actually is that 0 is divisible by all nos. So if k is 0, n is 0. And thus divisible by 6.

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by camitava » Wed Oct 03, 2007 7:01 am
Guys I am agree with Rateen and Kajcha! Stmt 1 is having some meaning and that is 100% correct. So IMO A. But B is not having any meaning at all. It is talking about only the value of K! How come n will be related to that?
Correct me If I am wrong


Regards,

Amitava