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

 It is currently 17 Aug 2018, 16:00

### 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
Your Progress

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 k is a positive integer and n = k(k + 7k), is n divisible

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Intern
Joined: 28 May 2013
Posts: 2
Location: United States
Concentration: Marketing, Healthcare
Schools: Duke '17 (WA)
WE: Military Officer (Military & Defense)
If k is a positive integer and n = k(k + 7k), is n divisible  [#permalink]

### Show Tags

Updated on: 10 Apr 2014, 06:42
8
46
00:00

Difficulty:

55% (hard)

Question Stats:

64% (01:11) correct 36% (01:20) wrong based on 1367 sessions

### HideShow timer Statistics

If k is a positive integer and n = k(k + 7), is n divisible by 6?

(1) k is odd.

(2) When k is divided by 3, the remainder is 2.

Originally posted by Jem2905 on 02 Nov 2013, 14:01.
Last edited by AbhiJ on 10 Apr 2014, 06:42, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
##### Most Helpful Expert Reply
Math Expert
Joined: 02 Sep 2009
Posts: 47977
Re: If k is a positive integer and n = k(k+7).....  [#permalink]

### Show Tags

10 Nov 2013, 12:30
20
17
ashsim wrote:
Chiranjeevee wrote:
Jem2905 wrote:
Hi guys, trying to get a little help on this problem that stumped me recently on a practice test. After going back and spending some more time with it, I got a different answer but I'm not sure if it's the right answer, and I'm not exactly sure I understand why it's the correct answer... any resphrasing of the question or statements will be hugely appreciated. Thanks!!

If k is a positive integer and n = k(k + 7k), is n divisible by 6?

(1) k is odd.

(2) When k is divided by 3, the remainder is 2.

Given, n= k(k+7K) = 8k^2
now for n to be divisible by 6, k should be divisible by 3.

1. K is odd.
clearly insufficient, for k=1 answer is No.
for k=3, answer is yes.

2. When k is divided by 3, the remainder is 2.
Remainder is 2, so K can never be divisible by 3,
Hence n will not be divisible by 6. So Sufficient.

IMO, B

Hi,
I'm trying to understand this question too. The question I saw had n=K(K+7), not k+ 7K as written in the question above. Can anyone explain this q with the change please?

Thanks!

If k is a positive integer and n = k(k + 7), is n divisible by 6?

(1) k is odd. If $$k = 1$$, then $$n = k(k + 7) = 8$$ and n is NOT divisible by 6 but if $$k = 3$$, then $$n = k(k + 7) = 30$$ and n IS divisible by 6. Not sufficient.

(2) When k is divided by 3, the remainder is 2 --> $$k = 3x + 2$$ --> $$n = k(k + 7) = (3x + 2)(3x + 9)=9x^2+33 x+18=3(3x^2+11x)+18$$. Notice that $$3x^2+11x$$ is even no matter whether x is even or odd, thus $$n=3(3x^2+11x)+18=3*even+(a \ multiple \ of \ 6)=(a \ multiple \ of \ 6)+(a \ multiple \ of \ 6)=(a \ multiple \ of \ 6)$$. Sufficient.

Answer: B.

Hope it's clear.
_________________
##### Most Helpful Community Reply
Manager
Joined: 15 Aug 2013
Posts: 57
Re: If k is a positive integer and n = k(k+7).....  [#permalink]

### Show Tags

11 Nov 2013, 03:23
12
2
Here is my solution-
If n=K(K+7), then-

1) k is odd => odd * (odd + Even) = Odd * Odd. We can not say it is multiple of 6 or not. Insufficient.
2) k = 3m +2. So, n=K(K+7) => n= (3m+2)(3m+9)= 3(3m+2)(m+3) => multiple of 3.
If m is odd, m+3 is even. Hence , multiple of 2. Also it is a multiple of 3 => multiple of 6
If m is even, 3m+2 is even. Hence , multiple of 2. Also it is a multiple of 3 => multiple of 6
So B is sufficient.
##### General Discussion
Intern
Joined: 28 Jan 2013
Posts: 31
Re: If k is a positive integer and n = k(k+7).....  [#permalink]

### Show Tags

02 Nov 2013, 22:08
2
3
Jem2905 wrote:
Hi guys, trying to get a little help on this problem that stumped me recently on a practice test. After going back and spending some more time with it, I got a different answer but I'm not sure if it's the right answer, and I'm not exactly sure I understand why it's the correct answer... any resphrasing of the question or statements will be hugely appreciated. Thanks!!

If k is a positive integer and n = k(k + 7k), is n divisible by 6?

(1) k is odd.

(2) When k is divided by 3, the remainder is 2.

Given, n= k(k+7K) = 8k^2
now for n to be divisible by 6, k should be divisible by 3.

1. K is odd.
clearly insufficient, for k=1 answer is No.
for k=3, answer is yes.

2. When k is divided by 3, the remainder is 2.
Remainder is 2, so K can never be divisible by 3,
Hence n will not be divisible by 6. So Sufficient.

IMO, B
Intern
Joined: 04 Nov 2013
Posts: 1
Re: If k is a positive integer and n = k(k+7).....  [#permalink]

### Show Tags

04 Nov 2013, 17:51
5
What about 5? It is a positive integer, when divided by 3 the remainder is 2 and 5(5+7)= 5(12) = 60, divisible by 6.
Intern
Joined: 09 Nov 2013
Posts: 10
Re: If k is a positive integer and n = k(k+7).....  [#permalink]

### Show Tags

10 Nov 2013, 08:45
Chiranjeevee wrote:
Jem2905 wrote:
Hi guys, trying to get a little help on this problem that stumped me recently on a practice test. After going back and spending some more time with it, I got a different answer but I'm not sure if it's the right answer, and I'm not exactly sure I understand why it's the correct answer... any resphrasing of the question or statements will be hugely appreciated. Thanks!!

If k is a positive integer and n = k(k + 7k), is n divisible by 6?

(1) k is odd.

(2) When k is divided by 3, the remainder is 2.

Given, n= k(k+7K) = 8k^2
now for n to be divisible by 6, k should be divisible by 3.

1. K is odd.
clearly insufficient, for k=1 answer is No.
for k=3, answer is yes.

2. When k is divided by 3, the remainder is 2.
Remainder is 2, so K can never be divisible by 3,
Hence n will not be divisible by 6. So Sufficient.

IMO, B

Hi,
I'm trying to understand this question too. The question I saw had n=K(K+7), not k+ 7K as written in the question above. Can anyone explain this q with the change please?

Thanks!
Intern
Joined: 27 Jun 2013
Posts: 3
Re: If k is a positive integer and n = k(k+7).....  [#permalink]

### Show Tags

07 Jan 2014, 15:01
Hey guys,

I still have a question on this problem, can someone explain why my solution is incorrect?

If k is a positive integer and n = k(k + 7), is n divisible by 6?

1. K is odd

Test cases:
K=1 n=1(1+7) = 8, is 8/6 NO
K=3 n=3(3+7)=30, is 30/6 YES
INSUFFICIENT

2. When k is divided by 3, the remainder is 2

Test Cases:
K=1 1/3=0 remainder 2, n=1(1+7) = 8, is 8/6 NO
K=5 5/3 =1 remainder 2, n=5(5+7) =60, is 60/6 YES
INSUFFICIENT

Combined:
K=1 overlaps - NO
K=5 overlaps - YES

I see the math approach in the posts above but why would the test cases produce a different result? What am I missing here?

Thanks in advance.
Math Expert
Joined: 02 Sep 2009
Posts: 47977
Re: If k is a positive integer and n = k(k+7).....  [#permalink]

### Show Tags

08 Jan 2014, 03:49
msbandi4321 wrote:
Hey guys,

I still have a question on this problem, can someone explain why my solution is incorrect?

If k is a positive integer and n = k(k + 7), is n divisible by 6?

1. K is odd

Test cases:
K=1 n=1(1+7) = 8, is 8/6 NO
K=3 n=3(3+7)=30, is 30/6 YES
INSUFFICIENT

2. When k is divided by 3, the remainder is 2

Test Cases:
K=1 1/3=0 remainder 2, n=1(1+7) = 8, is 8/6 NO
K=5 5/3 =1 remainder 2, n=5(5+7) =60, is 60/6 YES
INSUFFICIENT

Combined:
K=1 overlaps - NO
K=5 overlaps - YES

I see the math approach in the posts above but why would the test cases produce a different result? What am I missing here?

Thanks in advance.

1 divided by 3 yields the remainder of 1, not 2: 1=0*3+1.

Does this make sense?
_________________
Intern
Joined: 19 Nov 2013
Posts: 13
Schools: Haas '17 (S)
GMAT 1: 680 Q47 V36
Re: If k is a positive integer and n = k(k + 7k), is n divisible  [#permalink]

### Show Tags

09 Apr 2014, 16:51
I still don't see why Statement 2 is Sufficient. If you use 5, the outcome is 60 (divisible by 60), and if you use 8 the outcome is 320 (not divisible by 6). Please explain.
Math Expert
Joined: 02 Sep 2009
Posts: 47977
Re: If k is a positive integer and n = k(k + 7k), is n divisible  [#permalink]

### Show Tags

10 Apr 2014, 02:22
jbartuccio wrote:
I still don't see why Statement 2 is Sufficient. If you use 5, the outcome is 60 (divisible by 60), and if you use 8 the outcome is 320 (not divisible by 6). Please explain.

If k=8, then n=k(k+7)=8*15=120, not 320 and 120 is divisible by 6.

The reason why the second statement is sufficient is given here: if-k-is-a-positive-integer-and-n-k-k-7k-is-n-divisible-162594.html#p1290699 Does it make sense?
_________________
Intern
Joined: 08 Apr 2014
Posts: 12
Re: If k is a positive integer and n = k(k + 7k), is n divisible  [#permalink]

### Show Tags

10 Apr 2014, 02:58
jbartuccio wrote:
I still don't see why Statement 2 is Sufficient. If you use 5, the outcome is 60 (divisible by 60), and if you use 8 the outcome is 320 (not divisible by 6). Please explain.

if you refer to the initial question:
n= k(k+7k) = 8k^2

Hence, if k = 5, n = 8*5*5
if k = 8, n = 8*8*8

both of them are clearly not divisible by 6.

*press kudos if you like the answer
Intern
Joined: 08 Jan 2015
Posts: 2
Re: If k is a positive integer and n = k(k + 7k), is n divisible  [#permalink]

### Show Tags

02 Feb 2015, 19:16
1
1. K is odd

It's clearly insufficient. Test k=1 and k=3.

2. When k is divided by 3, the remainder is 2

n=k(k+7)=k(k+1+6) --> n=k(k+1)+6k
6k is always divisible by 6. Need to show that k(k+1) is also divisible by 6.

When k is divided by 3, the remainder is 2, then k=2,5,8,11,...

for k=2,5,8,11,..., k(k+1) is always divisible by 6:
k=2: k(k+1)=2*3=6
k=5: 5*6
k=8: 8*9
k=11: 11*12

k(k+1) and 6k are both divisible by 6, therefore n=k(k+1)+6k divisible by 6. sufficient

The answer is B.
Manager
Joined: 24 Dec 2014
Posts: 129
Location: India
Concentration: International Business, Finance
GMAT 1: 700 Q47 V39
GMAT 2: 790 Q51 V50
GPA: 3.65
WE: Operations (Energy and Utilities)
Re: If k is a positive integer and n = k(k + 7k), is n divisible  [#permalink]

### Show Tags

02 Feb 2015, 19:38
ricsingh wrote:
jbartuccio wrote:
I still don't see why Statement 2 is Sufficient. If you use 5, the outcome is 60 (divisible by 60), and if you use 8 the outcome is 320 (not divisible by 6). Please explain.

if you refer to the initial question:
n= k(k+7k) = 8k^2

Hence, if k = 5, n = 8*5*5
if k = 8, n = 8*8*8

both of them are clearly not divisible by 6.

*press kudos if you like the answer

Hi ricsingh:

n=k(k+7) is not same as n=k(k+7k).

Watch out for silly mistakes!

Posted from my mobile device
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8188
Location: Pune, India
Re: If k is a positive integer and n = k(k + 7k), is n divisible  [#permalink]

### Show Tags

02 Feb 2015, 22:29
5
3
Jem2905 wrote:
If k is a positive integer and n = k(k + 7), is n divisible by 6?

(1) k is odd.

(2) When k is divided by 3, the remainder is 2.

Given: n = k(k + 7)
Question: Is n divisible by 6?

(1) k is odd.
If k = 1, n = 8 - Not divisible by 6
If k = 6, n is divisible by 6
Not sufficient

(2) When k is divided by 3, the remainder is 2.
k = (3b+2)
n = (3b+2)(3b+2 + 7) = (3b + 2)(3b + 9) = 3*(3b + 2)(b + 3)
For n to be divisible by 6, it must be divisible by both 2 and 3. We see that it is divisible by 3. Let's see if it is divisible by 2 too i.e. if it is even.
b can be odd or even in this expression. If it is odd, (b+3) will become even because (Odd + Odd = Even). If it is even, (3b+2) will become even because (Even + Even = Even). So in either case, n will be even. So n will be divisible by 3 as well as 2 i.e. it will be divisible by 6.
Sufficient alone.

Answer (B)
_________________

Karishma
Veritas Prep GMAT Instructor

Save up to $1,000 on GMAT prep through 8/20! Learn more here > GMAT self-study has never been more personalized or more fun. Try ORION Free! Manager Joined: 08 Oct 2013 Posts: 51 Re: If k is a positive integer and n = k(k + 7k), is n divisible [#permalink] ### Show Tags 05 May 2015, 11:03 n= k(k+7) so is n divisible by 6? State 1: K is Odd Possible values are 1, 3, 5, 7 etc Expression is not true if n = 1 but true for rest of the numbers so Insufficient State 2: When K is divided by 3, the remainder is 2. Possible values are 2, 8, 11, 14, 17 etc For all there values k(k+7) is divisible by 6 - Hence State 2 is Sufficient. Answer is B Intern Joined: 08 Oct 2015 Posts: 10 Re: If k is a positive integer and n = k(k + 7k), is n divisible [#permalink] ### Show Tags 10 Nov 2015, 22:22 Bunuel wrote: If k is a positive integer and n = k(k + 7), is n divisible by 6? (1) k is odd. If $$k = 1$$, then $$n = k(k + 7) = 8$$ and n is NOT divisible by 6 but if $$k = 3$$, then $$n = k(k + 7) = 30$$ and n IS divisible by 6. Not sufficient. (2) When k is divided by 3, the remainder is 2 --> $$k = 3x + 2$$ --> $$n = k(k + 7) = (3x + 2)(3x + 9)=9x^2+33 x+18=3(3x^2+11x)+18$$. Notice that $$3x^2+11x$$ is even no matter whether x is even or odd, thus $$n=3(3x^2+11x)+18=3*even+(a \ multiple \ of \ 6)=(a \ multiple \ of \ 6)+(a \ multiple \ of \ 6)=(a \ multiple \ of \ 6)$$. Sufficient. Answer: B. Hope it's clear. Bunuel, Can you explain why : $$3x^2+11x$$ is even no matter whether x is even or odd? I'm sure there is a simple theoretical way to see this quicker than plugging in numbers. Thanks! Math Expert Joined: 02 Sep 2009 Posts: 47977 Re: If k is a positive integer and n = k(k + 7k), is n divisible [#permalink] ### Show Tags 10 Nov 2015, 23:35 dubyap wrote: Bunuel wrote: If k is a positive integer and n = k(k + 7), is n divisible by 6? (1) k is odd. If $$k = 1$$, then $$n = k(k + 7) = 8$$ and n is NOT divisible by 6 but if $$k = 3$$, then $$n = k(k + 7) = 30$$ and n IS divisible by 6. Not sufficient. (2) When k is divided by 3, the remainder is 2 --> $$k = 3x + 2$$ --> $$n = k(k + 7) = (3x + 2)(3x + 9)=9x^2+33 x+18=3(3x^2+11x)+18$$. Notice that $$3x^2+11x$$ is even no matter whether x is even or odd, thus $$n=3(3x^2+11x)+18=3*even+(a \ multiple \ of \ 6)=(a \ multiple \ of \ 6)+(a \ multiple \ of \ 6)=(a \ multiple \ of \ 6)$$. Sufficient. Answer: B. Hope it's clear. Bunuel, Can you explain why : $$3x^2+11x$$ is even no matter whether x is even or odd? I'm sure there is a simple theoretical way to see this quicker than plugging in numbers. Thanks! $$3x^2+11x=x(3x+11)$$. If x is even the result is obviously even: $$x(3x+11)=even*integer=even$$; If x is odd, then $$x(3x+11)=odd(odd*odd+odd)=odd*even=even$$. Hope it's clear. _________________ Intern Joined: 21 Jun 2014 Posts: 34 If k is a positive integer and n = k(k + 7k), is n divisible [#permalink] ### Show Tags 11 Nov 2015, 08:30 1 Jem2905 wrote: If k is a positive integer and n = k(k + 7), is n divisible by 6? (1) k is odd. (2) When k is divided by 3, the remainder is 2. Divisibility by 6 means must be divisible by 3x2 (Prime Factors of 6) 1) k is Odd ==> k(k+7) = Odd(Odd+Odd) = Even This will give me at least one factor of 2.. but nothing about factor of 3 [Insufficient] 2) k Divided by 3 ==> Remainder = 2 Now Possible values could be 2,5,8,11 -- Alternatively Even and Odd So if k=2,8,... Then I am getting One factor of 2 and when I add these to 7 I get Multiple of 3, Similarly when I I choose k=5,11.. Odd Number with remainder of 2 and I add them to 7 I get Even Multiple of 3 i.e. Multiple of 6 This is happening because, Numbers with Remainders of 2 when added to 7, which leaves remainder of 1 when divided by 3, makes the Sum Divisible by 3 [Sufficient] <--- Answer is B Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6028 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: If k is a positive integer and n = k(k + 7k), is n divisible [#permalink] ### Show Tags 15 Nov 2015, 10:43 Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. If k is a positive integer and n = k(k + 7), is n divisible by 6? (1) k is odd. (2) When k is divided by 3, the remainder is 2. There are 2 variables (n,k) and 1 equation (n=k(k+7)) in the original condition, 2 more equations in the given conditions, so there is high chance (D) will be our answer. For condition 1, the answer is 'no' for k=1, but 'yes' for k=5, so this is insufficient. For condition 2, k=3m+2 (m is a positive integer), or k+7=3m+2+7=3(m+3)=a multiple of 3 and k(k+7). This is always even and a multiple of 3, which means its always divisible by 6. This answers the question 'yes'; this is sufficient, and the answer is B. For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C 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$99 for 3 month Online Course"
"Free Resources-30 day online access & Diagnostic Test"
"Unlimited Access to over 120 free video lessons - try it yourself"

Senior Manager
Joined: 08 Dec 2015
Posts: 302
GMAT 1: 600 Q44 V27
Re: If k is a positive integer and n = k(k + 7k), is n divisible  [#permalink]

### Show Tags

27 Mar 2016, 09:06
VeritasPrepKarishma

Hello Karishma! A question here.. how do you quickly pull the 3 out of the initial statement?
(3b + 2)(3b + 9) = this one-> 3*(3b + 2)(b + 3)

Thank you!
Re: If k is a positive integer and n = k(k + 7k), is n divisible &nbs [#permalink] 27 Mar 2016, 09:06

Go to page    1   2    Next  [ 30 posts ]

Display posts from previous: Sort by

# If k is a positive integer and n = k(k + 7k), is n divisible

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

# Events & Promotions

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.