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

It is currently 21 Nov 2018, 06:06

LIVE NOW!

LBS is Calling R1 Admits - Join Chat Room to Catch the Latest Action


Close

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel
Events & Promotions in November
PrevNext
SuMoTuWeThFrSa
28293031123
45678910
11121314151617
18192021222324
2526272829301
Open Detailed Calendar
  • All GMAT Club Tests are Free and open on November 22nd in celebration of Thanksgiving Day!

     November 22, 2018

     November 22, 2018

     10:00 PM PST

     11:00 PM PST

    Mark your calendars - All GMAT Club Tests are free and open November 22nd to celebrate Thanksgiving Day! Access will be available from 0:01 AM to 11:59 PM, Pacific Time (USA)
  • Key Strategies to Master GMAT SC

     November 24, 2018

     November 24, 2018

     07:00 AM PST

     09:00 AM PST

    Attend this webinar to learn how to leverage Meaning and Logic to solve the most challenging Sentence Correction Questions.

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

Veritas Prep GMAT Instructor
User avatar
P
Joined: 16 Oct 2010
Posts: 8580
Location: Pune, India
Re: If k is a positive integer and n = k(k + 7k), is n divisible  [#permalink]

Show Tags

New post 29 Mar 2016, 21:34
1
iliavko wrote:
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!


Both terms of (3b + 9) have 3 as a factor.
(3*b + 3*3)

So you pull out a 3 from the brackets.
3*(b + 3)

Initial expression becomes
(3b + 2) * 3 * (b + 3)

The three terms are multiplied so you can arrange them in any way you like.

3 * (3b + 2) * (b + 3)
or
(b + 3) * (3b + 2) * 3
_________________

Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Intern
Intern
avatar
Joined: 08 Feb 2015
Posts: 4
Premium Member
Re: If k is a positive integer and n = k(k + 7k), is n divisible  [#permalink]

Show Tags

New post 16 Jun 2016, 13:09
sandeepmanocha wrote:
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



Agreed with Sandeep, if K is divisble by 3 and the remainder is 2, then (k+7) would divide evenly by 3 as 7/3 has a remainder of 1.
So 7/3=2*3+1 & k/3=a*3+2 -> (k+7)/3=(a*3+2)+(2*3+1)=3a+3+2*3=3(a+1+2) ->this is divisble by 3
Now, if k is odd k(k+7) is even, and if k is even k(k+7) is even, which means it is divisible by 2. Therefore it is divisible by both 2 and 3.
Manager
Manager
User avatar
B
Joined: 20 Jan 2017
Posts: 60
Location: United States (NY)
Schools: CBS '20 (A)
GMAT 1: 750 Q48 V44
GMAT 2: 610 Q34 V41
GPA: 3.92
Reviews Badge
Re: If k is a positive integer and n = k(k + 7k), is n divisible  [#permalink]

Show Tags

New post 29 Jan 2017, 15:34
1) First statement is insufficient. Smart numbers: k=1: 1*8=8, 8/6=1.666; k=3: 3*10=30, 30/6=5
2) To find out if n is divisible by 6 we need to find if n is divisible by 2 and divisible by 3. If k is even (e.g. 8, 14, 20...) then it is divisible by 2. If k is odd (5, 11, 17...), then k(odd)+7(odd)=even and it is divisible by 2. Since k/3 has a remainder of 2, and 7/3 has a remainder of 1, the remainders add up to another 3, thus k+7 is always divisible by 3.

Statement 2 alone is sufficient, while statement 1 alone is not sufficient.
Intern
Intern
avatar
Joined: 08 Jun 2011
Posts: 18
Re: If k is a positive integer and n = k(k + 7k), is n divisible  [#permalink]

Show Tags

New post 02 Jul 2017, 09:15
1
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 and 3. So, it is multiple of 6. Suff.
If m is even, 3m+2 is even. Hence , multiple of 2 and 3 again. So,multiple of 6. Suff.
Ans- B is
Manager
Manager
avatar
B
Joined: 23 Dec 2013
Posts: 159
Location: United States (CA)
GMAT 1: 710 Q45 V41
GMAT 2: 760 Q49 V44
GPA: 3.76
Reviews Badge
Re: If k is a positive integer and n = k(k + 7k), is n divisible  [#permalink]

Show Tags

New post 22 Jul 2017, 18:01
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.



The goal is to find if n is divisible by 2 and 3 (6).

Statement 1) K is odd.

K = 1
1(1+7) = 8, not divisible by 6

K = 3
3(3+7) = 30, divisible by 6.

Not sufficient.

Statement 2) K/3 R = 2.

If k/3 has a remainder of 2, then k could be equal to the following integers:

2 ==> 2(2+7) = 18, divisible by 6.

5 ==> 5(5+7) = 60, divisible by 6.

8 ==> 8(8+7) = 8*15 = 80+40 = 120, divisible by 6.

Sufficient.
Senior Manager
Senior Manager
avatar
S
Joined: 15 Jan 2017
Posts: 359
Re: If k is a positive integer and n = k(k + 7k), is n divisible  [#permalink]

Show Tags

New post 25 Sep 2017, 23:30
Got it in 5 minutes :(.
either way, St 1: n = k(k+7)means; plug in numbers.
if k is odd; then odd (odd + odd) --> odd.
if k is even; then even (even + odd) --> even.
Not suff

St 2: k/3 = Q + 2
5/3 = 1 + 2
8/3 = 2+ 2
So the lowest number completely divisible would be 5 (numerator: 3+2 and deno : 3+2). So 5/5 would be 1
Plugging K back to 5 --> 5(12) = 60 which is divisible by 6.
Thus B is sufficient.
Intern
Intern
User avatar
B
Joined: 16 Jul 2011
Posts: 41
Concentration: Marketing, Real Estate
GMAT 1: 550 Q37 V28
GMAT 2: 610 Q43 V31
Reviews Badge
Re: If k is a positive integer and n = k(k + 7k), is n divisible  [#permalink]

Show Tags

New post 05 Oct 2017, 06:12
VeritasPrepKarishma wrote:
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)

I suppose what I have marked in Red is a typo since the statement says k is odd.
_________________

"The fool didn't know it was impossible, so he did it."

Manager
Manager
User avatar
B
Joined: 05 Oct 2014
Posts: 54
Location: India
Concentration: General Management, Strategy
GMAT 1: 580 Q41 V28
GPA: 3.8
WE: Project Management (Energy and Utilities)
Re: If k is a positive integer and n = k(k + 7k), is n divisible  [#permalink]

Show Tags

New post 21 Oct 2017, 08:05
Hence n will not be divisible by 6. So Sufficient.

IMO, B[/quote]

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![/quote]

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.[/quote]



(1) Since k=odd, putting k = 3,5,7 we get n=30,60,98. 30&60 divisible but 98 is not divisible by 6. So INSUFFICIENT.

(2) says, k=3a+2, putting value of k in n = k(k+7), we get

n=(3a+2) * (3a+9) = 9a2+27a+6a+18

Now, (〖9a〗^2+27a+6a+18)/6 => 〖9a〗^2/6 + 27a/6 + 6a/6 + 18/6 => 〖9a〗^2/6 + 27a/6 + 0 + 0

When.

a = even, Reminder is 0

a = odd, Reminder is 1+1=2 [ a = 3,5,7 etc.]

Combining (1) & (2) , Reminder will be 2 as shown above. So Answer Choice should be ‘C’ .

Bunuel Can you please tell, where am I getting wrong ?
Intern
Intern
avatar
Joined: 24 Oct 2017
Posts: 2
Re: If k is a positive integer and n = k(k + 7k), is n divisible  [#permalink]

Show Tags

New post 24 Oct 2017, 07:58
I timed my response to work the problem--while knowing the answer--from the simplest solution I could find (quoted below) and it still took me ~3:30 mins/secs.

Any thoughts on how to cut down on the time used to solve? Has anyone else measured their response time?


KamiG wrote:
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
Manager
User avatar
B
Joined: 03 May 2017
Posts: 98
GMAT ToolKit User
If k is a positive integer and n = k(k + 7k), is n divisible  [#permalink]

Show Tags

New post 24 Oct 2017, 09:00
DanielAustin wrote:
I timed my response to work the problem--while knowing the answer--from the simplest solution I could find (quoted below) and it still took me ~3:30 mins/secs.

Any thoughts on how to cut down on the time used to solve? Has anyone else measured their response time?


KamiG wrote:
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.


Hi,

If you are having a problem, solving this, you should try to master the plugin method. Actually, I solved this in a minute without writing anything.

i.e Statement 1, says k=odd, then plug in the first two odd #s i.e 1 and 3.. and you had get no and yes respectively.. not sufficient

Statement 2, qualifies statement 1 as a special type of odd ie. \(3(Q) + 2 ... (5, 8, 11)\)...if you plug in any of this numbers you will get a yes, hence statement 2 alone is sufficient...

If you actually entered consecutive odd numbers in statement 1 you would see the pattern in statement 2 already. This is a reminder that both statements are usually equivalent from some perspective and you may be able to predict the answer this way.

Best,
Manager
Manager
User avatar
B
Joined: 14 Feb 2017
Posts: 51
Location: Australia
Concentration: Technology, Strategy
GMAT 1: 560 Q41 V26
GPA: 2.61
WE: Management Consulting (Consulting)
CAT Tests
Re: If k is a positive integer and n = k(k + 7k), is n divisible  [#permalink]

Show Tags

New post 14 Nov 2018, 14:16
TBH I didn't understand your approach Bunuel

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

So is n/6 = int?
(1) k is odd.
k=3
3(3+7)= 30
30/6 = int
=yes
k=1
1(8)= 8/6 = fraction
=no

Y/N are possible, so (1) --> IS

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

test the first few numbers that produce a remainder of 2 when divided by 3
2, 5, 8, 11
2(2+7) = 2(9) = 18
18/6 = int

5(12)/6 = int

8(15)/6
=4(15)/3
=4(5) (int)

11(18)/6 = int

More than enough to conclude this statement is Sufficient
GMAT Club Bot
Re: If k is a positive integer and n = k(k + 7k), is n divisible &nbs [#permalink] 14 Nov 2018, 14:16

Go to page   Previous    1   2   [ 31 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  


Copyright

GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

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®.