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

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If n and k are positive integers, is n divisible by 6? [#permalink]  27 Aug 2009, 16:15
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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
[Reveal] Spoiler: OA
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Re: Is n divisible by 6? [#permalink]  27 Aug 2009, 16:22
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1.) Sufficient

Always remember, when 3 consecutive integers are multiplied, then one should be a multiple of 3 and atleast one (or max 2) number is even. Also, if the multiple of 3 is also the even number like 6, 12, 18 then it is divisible by 6 always..

For ex:- 1*2*3 ; -4*-5*-6 ; 4*5*6 ; 11*12*13..

In any case; the product would be divisible by 6.

2.) insufficient.. does not say anything about n..

Ans. A
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Re: Is n divisible by 6? [#permalink]  27 Aug 2009, 20:47
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tarek99 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

Thanks

Since n = product of 3 consecutive numbers, it will be divisible by 3. We should also remember that 0 is also a multiple of 6. For eg : if k = 1,
n = 1*2*0 => n = 0 which IS divisible by 6.

I hope I am right. "0 is divisible by any number" right?
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Re: Is n divisible by 6? [#permalink]  27 Aug 2009, 22:50
(1) n = k(k + 1)(k - 1) --- Sufficient
(2) k – 1 is a multiple of 3 --- insufficient
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Re: Is n divisible by 6? [#permalink]  28 Aug 2009, 02:30
bhanushalinikhil wrote:
tarek99 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

Thanks

Since n = product of 3 consecutive numbers, it will be divisible by 3. We should also remember that 0 is also a multiple of 6. For eg : if k = 1,
n = 1*2*0 => n = 0 which IS divisible by 6.

I hope I am right. "0 is divisible by any number" right?

I slightly differ on your explination......specially if k = 1...

If k = 1 then n = 0*1*2 then n = 0 ... zero is not a positive integer hence all values of k will start from 2 onwards
hence values of n will start from 6 onwards.....but yes your ans is right A it is
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Re: Is n divisible by 6? [#permalink]  28 Aug 2009, 03:35
bhushan252 wrote:
bhanushalinikhil wrote:
tarek99 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

Thanks

Since n = product of 3 consecutive numbers, it will be divisible by 3. We should also remember that 0 is also a multiple of 6. For eg : if k = 1,
n = 1*2*0 => n = 0 which IS divisible by 6.

I hope I am right. "0 is divisible by any number" right?

I slightly differ on your explination......specially if k = 1...

If k = 1 then n = 0*1*2 then n = 0 ... zero is not a positive integer hence all values of k will start from 2 onwards
hence values of n will start from 6 onwards.....but yes your ans is right A it is

Right. Now see the mistake.
Thanks!
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Re: Is n divisible by 6? [#permalink]  15 Apr 2014, 06:54
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Re: If n and k are positive integers, is n divisible by 6? [#permalink]  24 Apr 2015, 15:17
Hello from the GMAT Club BumpBot!

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Re: If n and k are positive integers, is n divisible by 6? [#permalink]  16 May 2015, 03:15
Unsure how A is sufficient (I know that's the answer but I'm hoping someone could help show me why)

For me A is insufficient because:

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

we can get several cases, such as

k= 0
0*1*(-1) = -1 <-- not divisible by 6

k=1
1*2*0 = 0 <-- not divisible by 6

k=2
2*3*1 = 6 <--- divisible by 6

And so on. Statement 2 could supplement to my dilemma here, but since the answer is A, does anyone have a dumbed down explanation?
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Re: If n and k are positive integers, is n divisible by 6? [#permalink]  16 May 2015, 03:22
sabineodf wrote:
Unsure how A is sufficient (I know that's the answer but I'm hoping someone could help show me why)

For me A is insufficient because:

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

we can get several cases, such as

k= 0
0*1*(-1) = -1 <-- not divisible by 6

k=1
1*2*0 = 0 <-- not divisible by 6

k=2
2*3*1 = 6 <--- divisible by 6

And so on. Statement 2 could supplement to my dilemma here, but since the answer is A, does anyone have a dumbed down explanation?

JUST realised that the two first instances that I listed up are NOT positive integers hence they cannot be N. Ooops. Forget my question
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Re: If n and k are positive integers, is n divisible by 6? [#permalink]  16 May 2015, 03:54
Expert's post
sabineodf wrote:
Unsure how A is sufficient (I know that's the answer but I'm hoping someone could help show me why)

For me A is insufficient because:

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

we can get several cases, such as

k= 0
0*1*(-1) = -1 <-- not divisible by 6 = 0

k=1
1*2*0 = 0 <-- not divisible by 6

k=2
2*3*1 = 6 <--- divisible by 6

And so on. Statement 2 could supplement to my dilemma here, but since the answer is A, does anyone have a dumbed down explanation?

0 is divisible by every integer (except 0 itself).
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Re: If n and k are positive integers, is n divisible by 6? [#permalink]  16 May 2015, 05:18
Bunuel wrote:
sabineodf wrote:
Unsure how A is sufficient (I know that's the answer but I'm hoping someone could help show me why)

For me A is insufficient because:

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

we can get several cases, such as

k= 0
0*1*(-1) = -1 <-- not divisible by 6 = 0

k=1
1*2*0 = 0 <-- not divisible by 6

k=2
2*3*1 = 6 <--- divisible by 6

And so on. Statement 2 could supplement to my dilemma here, but since the answer is A, does anyone have a dumbed down explanation?

0 is divisible by every integer (except 0 itself).

Oh! Thanks, that´s good to know :D
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Re: If n and k are positive integers, is n divisible by 6? [#permalink]  23 May 2015, 01:54
tarek99 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

Here's my way to solve this:

Statement (1):
Rewrite n = k(k + 1)(k - 1) to: $$k^3-k$$ and plug in various positive integers to see that the result will alsways be divisible by 3. Therefore AC 1 = Sufficient

Statement (2)
Gives you just an idea about the term k-1, but nothing about k or k+1 itself. Therefore clearly insufficient.
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Re: If n and k are positive integers, is n divisible by 6?   [#permalink] 23 May 2015, 01:54
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