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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]  29 May 2008, 12:31
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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 is not equal to 1, is a multiple of 3.
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Re: DS - Integers [#permalink]  29 May 2008, 13:11
ah, good point. then it should be C
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Re: DS - Integers [#permalink]  29 May 2008, 16:04
Should be 'A'.
K can't be 1 as n is a positive int.
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Re: DS - Integers [#permalink]  29 May 2008, 16:07
kapilnegi wrote:
If n and k are positive integers, is n divisible by 6?

(1) n = k(k + 1)(k - 1)

(2) k is not equal to 1, is a multiple of 3.

i get D..

n= try any number..works..

even if k=1..n=0 which is divisible by 6..however n is positive so K cant be 1..

2) k=3M then K+1 will be an even number therefore N is divisible by 6..

whats the OA?
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Re: DS - Integers [#permalink]  29 May 2008, 19:28
Given : n and k are positive integers.

OA is C
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Re: DS - Integers [#permalink]  29 May 2008, 20:02
kapilnegi wrote:
Given : n and k are positive integers.

OA is C

i cant agree with the OA..

whats the source?
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Re: DS - Integers [#permalink]  29 May 2008, 20:27
I agree with the OA - the answer is C - both statements are required - for S1, if k is 1, then it doesn't hold true, S2 doesn't hold good by itself - so with S1 and S2 we can say that n is divisible by 6 - even with the statement that n is a positive integer. Any other logic?
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Re: DS - Integers [#permalink]  29 May 2008, 20:36
This is besides the point, but
Why in the world do you guys think if k=1, then n is not divisible by 6?
If k=1, then n=0
0 is divisible by any number.
Am I missing something here?

BTW, I got A. I don't agree with the OA.
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Re: DS - Integers [#permalink]  29 May 2008, 20:37
tharunv wrote:
I agree with the OA - the answer is C - both statements are required - for S1, if k is 1, then it doesn't hold true, S2 doesn't hold good by itself - so with S1 and S2 we can say that n is divisible by 6 - even with the statement that n is a positive integer. Any other logic?

if K=1 then n=0..for argument sake 0/6 is divisble by 6..
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Re: DS - Integers [#permalink]  29 May 2008, 20:45
If that is the logic, then 0 is divisible by all numbers - that is besides the point - I still say the OA is correct - otherwise it is a trial question meant to confuse people. Anyways we can continue with the argument. Quick question - 0 is not counted as a positive integer - if we say k not equal to 1, then we can conclude that S1 is sufficient
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Re: DS - Integers [#permalink]  29 May 2008, 21:37
kapilnegi wrote:
If n and k are positive integers, is n divisible by 6?

(1) n = k(k + 1)(k - 1)

(2) k is not equal to 1, is a multiple of 3.

What are u guys thinking
Stat 1 is clearly sufficient

n and k are positive integers

n and k cannot be zero

so in statement if u think of taking k as 1 then n=0 and 0 is not positive(see the line in red) So according to the given details k has to be greater than 1

so stat 1 is suff

now lets see that second stat

k is not equal to 1, is a multiple of 3

as every second number is even and every third number is a multiple of 3 this is also correct

so according to me ans is d
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Re: DS - Integers [#permalink]  30 May 2008, 03:56
rohit929: statement 2 doesn tell anything about n, how can it be suff. Remember when your are reading statement 2, forget about 1.

I'll go with A.
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Re: DS - Integers [#permalink]  30 May 2008, 04:17
I get A as well. Durgesh makes a good point that you cannot take anything you've learned in (1) when you process (2).

The only way I could see C working is if it said that n and k were non-negative integers.... Positive integers, by definition, cannot include zero.
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Re: DS - Integers [#permalink]  30 May 2008, 07:37
bkk145 wrote:
This is besides the point, but
Why in the world do you guys think if k=1, then n is not divisible by 6?
If k=1, then n=0
0 is divisible by any number.
Am I missing something here?

BTW, I got A. I don't agree with the OA.

This was my argument as well. This question sucks.
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Re: DS - Integers [#permalink]  31 May 2008, 06:04
VPRedSoxFan wrote:
I get A as well. Durgesh makes a good point that you cannot take anything you've learned in (1) when you process (2).

The only way I could see C working is if it said that n and k were non-negative integers.... Positive integers, by definition, cannot include zero.

Durgesh and VPRedSoxFan n and k are positive are given in th sten not in statement 1
If n and k are positive integers, is n divisible by 6?

(1) n = k(k + 1)(k - 1)

(2) k is not equal to 1, is a multiple of 3.
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Re: DS - Integers [#permalink]  31 May 2008, 11:10
kapilnegi wrote:
If n and k are positive integers, is n divisible by 6?

(1) n = k(k + 1)(k - 1)
(2) k is not equal to 1, "xxx" is a multiple of 3.

is something missing in st 2 after "k is not equal to 1," ?

from the question stem, it is clear that n and k are +ve.

1 says n is multiple of 3 +ve consecutive integer. any 3 +ve (in fact any -ve as well) consecutive integers are evenly divisible by 6. so n is divisible by 6. suff..

2 says k is an integer >1 but it has nothing to do with n. so insuff...

therefore, A.
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Re: DS - Integers   [#permalink] 31 May 2008, 11:10
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