GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 14 Dec 2019, 05:29

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

### Hide Tags

VP
Joined: 21 Jul 2006
Posts: 1011
If n and k are positive integers, is n divisible by 6?  [#permalink]

### Show Tags

27 Aug 2009, 17:15
10
71
00:00

Difficulty:

35% (medium)

Question Stats:

65% (01:14) correct 35% (01:38) wrong based on 820 sessions

### HideShow timer Statistics

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
Math Expert
Joined: 02 Sep 2009
Posts: 59724
Re: If n and k are positive integers, is n divisible by 6?  [#permalink]

### Show Tags

16 May 2015, 04:54
4
6
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).
_________________
Manager
Joined: 25 Aug 2009
Posts: 132
Re: Is n divisible by 6?  [#permalink]

### Show Tags

27 Aug 2009, 17:22
12
11
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
##### General Discussion
Manager
Joined: 28 Jul 2009
Posts: 85
Location: India
Schools: NUS, NTU, SMU, AGSM, Melbourne School of Business
Re: Is n divisible by 6?  [#permalink]

### Show Tags

27 Aug 2009, 21:47
8
2
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?
Manager
Joined: 18 Jul 2009
Posts: 131
Location: India
Schools: South Asian B-schools
Re: Is n divisible by 6?  [#permalink]

### Show Tags

28 Aug 2009, 03:30
2
2
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
Manager
Joined: 28 Jul 2009
Posts: 85
Location: India
Schools: NUS, NTU, SMU, AGSM, Melbourne School of Business
Re: Is n divisible by 6?  [#permalink]

### Show Tags

28 Aug 2009, 04:35
1
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!
Manager
Joined: 28 Jan 2015
Posts: 123
Concentration: General Management, Entrepreneurship
GMAT 1: 670 Q44 V38
Re: If n and k are positive integers, is n divisible by 6?  [#permalink]

### Show Tags

16 May 2015, 04:15
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?
Manager
Joined: 28 Jan 2015
Posts: 123
Concentration: General Management, Entrepreneurship
GMAT 1: 670 Q44 V38
Re: If n and k are positive integers, is n divisible by 6?  [#permalink]

### Show Tags

16 May 2015, 04:22
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
Manager
Joined: 28 Jan 2015
Posts: 123
Concentration: General Management, Entrepreneurship
GMAT 1: 670 Q44 V38
Re: If n and k are positive integers, is n divisible by 6?  [#permalink]

### Show Tags

16 May 2015, 06:18
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
Retired Moderator
Joined: 29 Apr 2015
Posts: 816
Location: Switzerland
Concentration: Economics, Finance
Schools: LBS MIF '19
WE: Asset Management (Investment Banking)
Re: If n and k are positive integers, is n divisible by 6?  [#permalink]

### Show Tags

23 May 2015, 02:54
1
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.
Intern
Joined: 21 Mar 2014
Posts: 21
If n and k are positive integers, is n divisible by 6?  [#permalink]

### Show Tags

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?
_________________
kinaare paaon phailane lage hian,
nadi se roz mitti kat rahi hai....
CEO
Joined: 20 Mar 2014
Posts: 2560
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: If n and k are positive integers, is n divisible by 6?  [#permalink]

### Show Tags

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.
CEO
Joined: 20 Mar 2014
Posts: 2560
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
If n and k are positive integers, is n divisible by 6?  [#permalink]

### Show Tags

19 Sep 2015, 06:01
1
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!).
Math Expert
Joined: 02 Aug 2009
Posts: 8318
Re: If n and k are positive integers, is n divisible by 6?  [#permalink]

### Show Tags

19 Sep 2015, 06:37
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..
_________________
CEO
Joined: 20 Mar 2014
Posts: 2560
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
If n and k are positive integers, is n divisible by 6?  [#permalink]

### Show Tags

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.
Math Expert
Joined: 02 Aug 2009
Posts: 8318
Re: If n and k are positive integers, is n divisible by 6?  [#permalink]

### Show Tags

19 Sep 2015, 06:43
1
Engr2012 wrote:
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 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...
_________________
Manager
Status: One Last Shot !!!
Joined: 04 May 2014
Posts: 229
Location: India
Concentration: Marketing, Social Entrepreneurship
GMAT 1: 630 Q44 V32
GMAT 2: 680 Q47 V35
Re: If n and k are positive integers, is n divisible by 6?  [#permalink]

### Show Tags

09 Nov 2015, 08:21
Bunuel wrote:

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

Hi Bunuel

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??
Math Expert
Joined: 02 Aug 2009
Posts: 8318
Re: If n and k are positive integers, is n divisible by 6?  [#permalink]

### Show Tags

09 Nov 2015, 08:31
arhumsid wrote:
Bunuel wrote:

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

Hi Bunuel

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??

hi ,
if you mean first n consecutive integers..
then product of n consecutive integers is nothing but n! itself...
_________________
Manager
Status: One Last Shot !!!
Joined: 04 May 2014
Posts: 229
Location: India
Concentration: Marketing, Social Entrepreneurship
GMAT 1: 630 Q44 V32
GMAT 2: 680 Q47 V35
Re: If n and k are positive integers, is n divisible by 6?  [#permalink]

### Show Tags

09 Nov 2015, 09:55
chetan2u wrote:
arhumsid wrote:
Bunuel wrote:

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

Hi Bunuel

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??

hi ,
if you mean first n consecutive integers..
then product of n consecutive integers is nothing but n! itself...

Nope i dont mean that.. I mean 'ANY n consecutive Integers'. What do you say about that?
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 8261
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: If n and k are positive integers, is n divisible by 6?  [#permalink]

### Show Tags

10 Nov 2015, 11:52
2
Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.

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

This is a recently common type of question.
There are 2 variables (n,k) and 2 equations are given from the 2 conditions, so there is high chance (C) will be our answer. Looking at the conditions individually,
for condition 1, n becomes a product of 3 consecutive integers, so this always makes it a multiple of 6, so this is sufficient.
for condition 2, there is not information about n, so this is insufficient, and the answer becomes (A).
We cannot look at the conditions together as this is a commonly made mistake type 4(A).

For cases where we need 2 more equation, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
_________________
MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
"Only \$79 for 1 month Online Course"
"Free Resources-30 day online access & Diagnostic Test"
"Unlimited Access to over 120 free video lessons - try it yourself"
Re: If n and k are positive integers, is n divisible by 6?   [#permalink] 10 Nov 2015, 11:52

Go to page    1   2    Next  [ 25 posts ]

Display posts from previous: Sort by