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If n and k are positive integers, is n divisible by 6? (1) n

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If n and k are positive integers, is n divisible by 6? (1) n [#permalink] New post 01 Oct 2007, 05:47
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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.
Director
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Re: DS - sets [#permalink] New post 01 Oct 2007, 06:53
Balvinder wrote:
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.


A. k (k + 1) (k - 1) are consequtive integers...
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Re: DS - sets [#permalink] New post 01 Oct 2007, 07:05
Balvinder wrote:
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.


St.1: Insuf. b/c K=???
St.2: Insuf.- nothing about N????
1&2: Suffic.
n=k(k^2-1)
k-1 is a mult. of 3-->( k^2-1) is a mult of 3.
k=10 n=10x(100-1)=10x99=990 -->990/3=330
k=7 n=7x(49-1)=7x48=336-->336/6=56
C.
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 [#permalink] New post 01 Oct 2007, 07:15
St1:

n = product of three consecutive integers.

If k = 1, then n = 0*1*2 = 0 ---> divisible by 6
If k = 2, then n = 1*2*3 = 6 ---> divisible by 6
If k = 3, then n = 3*4*5 = 60 ---> divisible by 6

Always divisible by 6. Sufficient.

St2:
Nothing about n. INsufficient.

Ans A
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 [#permalink] New post 01 Oct 2007, 10:57
A

Plug in numbers strategy. n is always divisible by 6 if stat 1 is true.
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Re: DS - sets [#permalink] New post 01 Oct 2007, 11:04
Balvinder wrote:
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.


A
1. use plug in method. (k-1)*k*(k+1)=N, this multiplication always has multiple of both 3 and 2 then the product is divisible by 6. suff

2. hmm. no info about N so out. insuff.
Re: DS - sets   [#permalink] 01 Oct 2007, 11:04
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