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n divisible by 6?

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by crackgmat007 » Wed May 13, 2009 9:08 pm
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.

Is statement 1 alone is sufficient? Can anyone clarify this pls? Tx much.
Last edited by crackgmat007 on Mon May 18, 2009 10:48 am, edited 1 time in total.
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Source: — Data Sufficiency |
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by aj5105 » Wed May 13, 2009 10:31 pm
Statement (1): You cannot consider 0,1,2. Though 1,2 are positive, the product (n) becomes 0.

Consider any three consecutive numbers[ k-1, k, k+1] such as 1, 2, 3 or 2, 3, 4 or 4, 5, 6 ... Their product is divisible by 6.

(A)
Last edited by aj5105 on Fri May 15, 2009 7:23 pm, edited 1 time in total.
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by mike22629 » Fri May 15, 2009 9:28 am
You can also consider k=1

0 is divisible by any number or 0 is a multiple of every number.

IMO B.

Furthermore, think of the equation in the first statement

It is telling you that n = the product of three consecutive numbers

A number is divisible by 6 is it is divisible by 3 and by 2. Three consecutive positive numbers will always meet this requirement.
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by iamcste » Fri May 15, 2009 1:41 pm
mike22629 wrote:You can also consider k=1

0 is divisible by any number or 0 is a multiple of every number.

IMO B.

Furthermore, think of the equation in the first statement

It is telling you that n = the product of three consecutive numbers

A number is divisible by 6 is it is divisible by 3 and by 2. Three consecutive positive numbers will always meet this requirement.
IMO AHow can it be B

In Stmt 2, we have only information only about K. How does it help you to determine divisibility of n by 6. means you considered information stated in stmt 1 without knowing :D
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by iamcste » Fri May 15, 2009 1:46 pm
mike22629 wrote:You can also consider k=1

0 is divisible by any number or 0 is a multiple of every number.

.
when you select K=1, n becomes 0...and as per info given, n is a positive integer...and hence it cant be 0, Note: 0 is neither positive nor negative

HTH
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