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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..

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

Please explain your answer. Thanks

My answer would be A. 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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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

Please explain your answer. Thanks

My answer would be A. 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 _________________

Bhushan S. If you like my post....Consider it for Kudos

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

Please explain your answer. Thanks

My answer would be A. 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: If n and k are positive integers, is n divisible by 6? [#permalink]

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24 Apr 2015, 16:17

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Re: If n and k are positive integers, is n divisible by 6? [#permalink]

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23 May 2015, 02:54

1

This post was BOOKMARKED

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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If n and k are positive integers, is n divisible by 6? [#permalink]

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19 Sep 2015, 05:17

i dont understand why b is not sufficient, if k-1 is a multiple of 3, then k or k+1 should be a multiple of 2... and so, it should be sufficient as well.... am i missing something?
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Re: If n and k are positive integers, is n divisible by 6? [#permalink]

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19 Sep 2015, 05:53

saroshgilani wrote:

i dont understand why b is not sufficient, if k-1 is a multiple of 3, then k or k+1 should be a multiple of 2... and so, it should be sufficient as well.... am i missing something?

You are correct to say that if k-1 =3p, then k or k+1 will be =2r. But what is the connection between k and n? You are using information from statement 1 to check for sufficiency of statement 2. This absolutely wrong. You need to isolate both statements completely to determine their respective sufficiency. Only once they are not sufficient ALONE, you go to combining the 2 statements.

If n and k are positive integers, is n divisible by 6? [#permalink]

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19 Sep 2015, 06:01

1

This post received KUDOS

reto 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

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.

reto, a couple of points about your solution (although you were able to get to the correct answer):

For statement 1, k(k-1)(k+1) will ALWAYS be divisible by 6. Think of this way. For any number to be divisible by 6, it needs to be divisible by BOTH 2 and 3 at the same time.

Consider 2 cases, if k =2p, then k+1 or k-1 MUST be disivible by 3. Thus the product k(k-1)(k+1) will always be divisible by 6.

Case 2: if k = 3p, then k-1 or k+1 MUST be divisble by 2. Thus the product k(k-1)(k+1) will always be divisible by 6.

As for your interpretation of statement 2, we do not need to even look at the statement and start computing numbers as the question stem asks us whether "n" is divisible by 6 and NOT whether "k" is divisible by 6. Statement 2 does not provide any relation between k and n and hence straightaway you can reject this statement.

Also you are incorrect to say that as k-1=3p, you have no information about k+1 or k. This is not correct. If k-1=3p., then 1 of k or k+1 MUST be divisible by 2 (you can check with some numbers). Thus you do get some actionable information (albeit not pertinent to the original question asked!).

Also you are incorrect to say that as k-1=3p, you have no information about k+1 or k. This is not correct. If k-1=3p., then 1 of k or k+1 MUST be divisible by 2 (you can check with some numbers). Thus you do get some actionable information (albeit not pertinent to the original question asked!).

Hi Engr2012, irrespective of what K-1 is, one of k or k+1 will always be divisible by 2 as k is a positive integer..
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If n and k are positive integers, is n divisible by 6? [#permalink]

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19 Sep 2015, 06:38

chetan2u wrote:

Engr2012 wrote:

Also you are incorrect to say that as k-1=3p, you have no information about k+1 or k. This is not correct. If k-1=3p., then 1 of k or k+1 MUST be divisible by 2 (you can check with some numbers). Thus you do get some actionable information (albeit not pertinent to the original question asked!).

Hi Engr2012, irrespective of what K-1 is, one of k or k+1 will always be divisible by 2 as k is a positive integer..

Yes that is what I have written above. Additionally, the question asks about n and NOT k. Without any relation between n and k , this information is of no use.

Also you are incorrect to say that as k-1=3p, you have no information about k+1 or k. This is not correct. If k-1=3p., then 1 of k or k+1 MUST be divisible by 2 (you can check with some numbers). Thus you do get some actionable information (albeit not pertinent to the original question asked!).

Hi Engr2012, irrespective of what K-1 is, one of k or k+1 will always be divisible by 2 as k is a positive integer..

Yes but which one will it be ? Additionally, the question asks about n and NOT k. Without any relation between n and k , this information is of no use.

Hi, I am not questioning the solution. it is very clear and fairly simple... i am just pointing to the highlighted portion that it is not because k-1=3p that k or k+1 will be div by 2 but one of k or k+1 will always be div by 2... _________________

I just realized something while solving this question, please let me know if im correct on this one:

Product of n consecutive integers is always divisible by -all integers between 1 and n inclusive. -n! -all the factors of n!

Can we make this generalization??
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