Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
GMAT Club

 It is currently 22 Mar 2017, 21:11

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# If n is a positive integer, and r is the remainder when

Author Message
TAGS:

### Hide Tags

Director
Joined: 03 Sep 2006
Posts: 878
Followers: 6

Kudos [?]: 826 [5] , given: 33

If n is a positive integer, and r is the remainder when [#permalink]

### Show Tags

27 Apr 2010, 09:45
5
KUDOS
15
This post was
BOOKMARKED
00:00

Difficulty:

35% (medium)

Question Stats:

72% (02:19) correct 28% (02:04) wrong based on 558 sessions

### HideShow timer Statistics

If n is a positive integer and r is the remainder when 4 + 7n is divided by 3, what is the value of r?

(1) n+1 is divisible by 3
(2) n>20.
[Reveal] Spoiler: OA

Last edited by Bunuel on 14 Mar 2012, 05:48, edited 1 time in total.
Edited the question and added the OA
Math Expert
Joined: 02 Sep 2009
Posts: 37547
Followers: 7390

Kudos [?]: 99198 [6] , given: 11008

### Show Tags

27 Apr 2010, 10:23
6
KUDOS
Expert's post
10
This post was
BOOKMARKED
LM wrote:

If n is a positive integer, and r is the remainder when 4 + 7n is divided by 3, what is the value of r?

r is the remainder when 4n+7 is divided by 3 --> $$4+7n=3q+r$$, where $$r$$ is an integer $$0\leq{r}<3$$. $$r=?$$

(1) n+1 is divisible by 3 --> $$n+1=3k$$, or $$n=3k-1$$ --> $$4+7(3k-1)=3q+r$$ --> $$3(7k-1-q)=r$$ --> so $$r$$ is multiple of 3, but it's an integer in the range $$0\leq{r}<3$$. Only multiple of 3 in this range is 0 --> $$r=0$$. Sufficient.

(2) n>20. Clearly not sufficient. $$n=21$$, $$4+7n=151=3q+r$$, $$r=1$$ BUT $$n=22$$, $$4+7n=158=3q+r$$, $$r=2$$. Not sufficient.

P.S. Please post DS questions in DS subforum.
_________________
Retired Moderator
Joined: 02 Sep 2010
Posts: 804
Location: London
Followers: 108

Kudos [?]: 985 [2] , given: 25

### Show Tags

30 Sep 2010, 22:37
2
KUDOS
2
This post was
BOOKMARKED
Michmax3 wrote:
If n is a positive integer and r is the remainder when 4+7n is divided by 3. What is the value of r?

1) n+1 is divisible by 3
2) n>20

7n+4=3(2n+1) + n+1

So remainder when divided by 3 will be same as remainder left by n+1

1) sufficient ... Gives the answer
2) insufficient ... Irrelevant. Eg n= 22,23,24 all are possible

_________________
Current Student
Joined: 07 May 2010
Posts: 731
Followers: 14

Kudos [?]: 88 [1] , given: 66

### Show Tags

27 Jul 2010, 16:21
1
KUDOS
i think i may have an easier way....
s1) 7n+4 = (6n+3)+(n+1)
if (n+1)/3 = an integer, so must 3 times (n+1)....which is (6n+3)
s2) Obviously NS
_________________
Intern
Joined: 27 Aug 2012
Posts: 16
Schools: Anderson '15
Followers: 0

Kudos [?]: 12 [1] , given: 8

If n is a positive integer, and r is the remainder when 4 + 7n i [#permalink]

### Show Tags

26 Oct 2012, 05:47
1
KUDOS
A.

1) n+1 = 3 X (X is your Quotient) + 0(Remainder)
n+1=3X
Any Multiple of N+1 will be divisible by 3
Multiply by 7 --> 7(n+1)
So 7n+7 is divisible by 3 , implies 7n+4 is divisible by 3.

try nos for verification

2) n> 20 --- NS
Director
Joined: 03 Sep 2006
Posts: 878
Followers: 6

Kudos [?]: 826 [0], given: 33

### Show Tags

27 Apr 2010, 10:29
Bunuel wrote:
LM wrote:

If n is an integer, and r is the remainder when 4+7n is divided by 3, what is the value of r?

r is the remainder when 4n+7 is divided by 3 --> $$4+7n=3q+r$$, where $$r$$ is an integer $$0\leq{r}<3$$. $$r=?$$

(1) n+1 is divisible by 3 --> $$n+1=3k$$, or $$n=3k-1$$ --> $$4+7(3k-1)=3q+r$$ --> $$3(7k-1-p)=r$$ --> so $$r$$ is multiple of 3, but it's an integer in the range $$0\leq{r}<3$$. Only multiple of 3 in this range is 0 --> $$r=0$$. Sufficient.

(2) n>20. Clearly not sufficient. $$n=21$$, $$4+7n=151=3q+r$$, $$r=1$$ BUT $$n=22$$, $$4+7n=158=3q+r$$, $$r=2$$. Not sufficient.

P.S. Please post DS questions in DS subforum.

Thanks for detailed explanation. Sorry...in a haste I posted the DS questions in different forum. I will rectify this mistake next time onwards. Exam date is nearby ..so was in haste to get the explanations...thanks again..

You mentioned the $$0\leq{r}<3$$ above, because remainder can't be more than the divisor. Is this correct?
Math Expert
Joined: 02 Sep 2009
Posts: 37547
Followers: 7390

Kudos [?]: 99198 [0], given: 11008

### Show Tags

29 Apr 2010, 22:47
LM wrote:

You mentioned the $$0\leq{r}<3$$ above, because remainder can't be more than the divisor. Is this correct?

Yes, you are right.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 37547
Followers: 7390

Kudos [?]: 99198 [0], given: 11008

### Show Tags

27 Jul 2010, 16:40
Expert's post
2
This post was
BOOKMARKED
omarjmh wrote:
i think i may have an easier way....
s1) 7n+4 = (6n+3)+(n+1)
if (n+1)/3 = an integer, so must 3 times (n+1)....which is (6n+3)
s2) Obviously NS

Little correction: 3 times (n+1) is 3n+3 not (6n+3).

But you are right, we can solve with this approach as well:

(1) n+1 is divisible by 3 --> $$7n+4=(4n+4)+3n=4(n+1)+3n$$ --> $$4(n+1)$$ is divisible by 3 as $$n+1$$ is, and $$3n$$ is obviously divisible by 3 as it has 3 as multiple, thus their sum, $$7n+4$$, is also divisible by 3, which means that remainder upon division $$7n+4$$ by 3 will be 0. Sufficient.

Hope it's clear.
_________________
Senior Manager
Status: Current Student
Joined: 14 Oct 2009
Posts: 371
Schools: Chicago Booth 2013, Ross, Duke , Kellogg , Stanford, Haas
Followers: 13

Kudos [?]: 114 [0], given: 53

### Show Tags

30 Sep 2010, 22:45
shrouded1 wrote:
Michmax3 wrote:
7n+4=3(2n+1) + n+1

So remainder when divided by 3 will be same as remainder left by n+1

Can you explain how you get to 7n+4=3(2n+1)?
_________________
Senior Manager
Status: Time to step up the tempo
Joined: 24 Jun 2010
Posts: 408
Location: Milky way
Schools: ISB, Tepper - CMU, Chicago Booth, LSB
Followers: 8

Kudos [?]: 203 [0], given: 50

### Show Tags

30 Sep 2010, 22:51
The information in the statement A is used favorably to tweak the equation in the question.

Hence 7n+4 becomes 3(2n+1) + (n+1). Now since 3(2n+1) leaves a remainder 0 when divided by 3, the remainder of 7n+4 will be the same as the remainder of (n+1).

Since (n+1) is also given in option A to be divisible by 3, hence remainder 0. Statement A is sufficient.
_________________

Support GMAT Club by putting a GMAT Club badge on your blog

Retired Moderator
Joined: 02 Sep 2010
Posts: 804
Location: London
Followers: 108

Kudos [?]: 985 [0], given: 25

### Show Tags

30 Sep 2010, 22:56
Michmax3 wrote:
Can you explain how you get to 7n+4=3(2n+1)?

Just trying to split it out into parts divisible by 3
7n becomes 6n+n
4 becomes 3+1
_________________
Senior Manager
Status: Current Student
Joined: 14 Oct 2009
Posts: 371
Schools: Chicago Booth 2013, Ross, Duke , Kellogg , Stanford, Haas
Followers: 13

Kudos [?]: 114 [0], given: 53

### Show Tags

01 Oct 2010, 10:25
shrouded1 wrote:
Michmax3 wrote:
Can you explain how you get to 7n+4=3(2n+1)?

Just trying to split it out into parts divisible by 3
7n becomes 6n+n
4 becomes 3+1

Thanks I see it now...btw is your avatar from the opening credits of Dexter?
_________________
Retired Moderator
Joined: 02 Sep 2010
Posts: 804
Location: London
Followers: 108

Kudos [?]: 985 [0], given: 25

### Show Tags

01 Oct 2010, 10:36
Michmax3 wrote:
shrouded1 wrote:
Michmax3 wrote:
Can you explain how you get to 7n+4=3(2n+1)?

Just trying to split it out into parts divisible by 3
7n becomes 6n+n
4 becomes 3+1

Thanks I see it now...btw is your avatar from the opening credits of Dexter?

YES !!

It took me a lot of time to erase the credits which were in deep red painted right across the face and still maintain a semblance of originality in the image ....
_________________
Retired Moderator
Joined: 02 Sep 2010
Posts: 804
Location: London
Followers: 108

Kudos [?]: 985 [0], given: 25

### Show Tags

01 Oct 2010, 10:50
I even replaced my facebook photo with this one
_________________
Manager
Joined: 11 Jul 2010
Posts: 225
Followers: 3

Kudos [?]: 91 [0], given: 20

### Show Tags

01 Oct 2010, 23:26
Profound identity question:

isnt "ezhilkumarank" the one with the dexter image?!! and not shrouded...
Retired Moderator
Joined: 02 Sep 2010
Posts: 804
Location: London
Followers: 108

Kudos [?]: 985 [0], given: 25

### Show Tags

02 Oct 2010, 00:05
there is the cartoon dexter .... and then there is the serial killer dexter

very different things !
_________________
Manager
Joined: 20 Jul 2010
Posts: 52
Followers: 3

Kudos [?]: 60 [0], given: 51

### Show Tags

20 Oct 2010, 23:26
Why am i getting negative Reminder here?

Given: 4 + 7n = 3q + r

If we simplify this further,

7n = 3q + r -4
n = (3q/7) + (r-4)/7
n + 1 = (3q/7) + (r-4)/7 + 1 ----- (1)

But From Statement 1, n+1 is divisible by 3.
So, (1) is divisible by 3 and hence the reminder is Zero
i.e., (r-4/7) + 1 = 0 ==> r = -3

Cheers!
Ravi
Senior Manager
Status: mba here i come!
Joined: 07 Aug 2011
Posts: 270
Followers: 44

Kudos [?]: 1084 [0], given: 48

### Show Tags

14 Mar 2012, 05:31
1) n+1 = 3k and 7(n+1) = 7*3k
so, 7n+7 is divisible by 3. this means that 7n+4 must also be divisible by 3.
_________________

press +1 Kudos to appreciate posts

Director
Joined: 03 Aug 2012
Posts: 915
Concentration: General Management, General Management
GMAT 1: 630 Q47 V29
GMAT 2: 680 Q50 V32
GPA: 3.7
WE: Information Technology (Investment Banking)
Followers: 23

Kudos [?]: 734 [0], given: 322

Re: If n is a positive integer, and r is the remainder when [#permalink]

### Show Tags

09 Aug 2013, 09:17
Q. What is r?

(1). (n+1) div by 3

By question stem we know that

4 + 7n = 3Q + r

Splitting the equation as below

4+4n+3n = 3Q + r

=> 3n + 4 (n+1) = 3Q + r

LHS is divisible by 3 as (n+1) is div by 3, so RHS should also be divisible by 3 hence r should be 0

(2).

4 + 7n = 3Q +r

Case 1: n=21

4 + 7*21 = 3Q +r

LHS gives 4 as remainder when divided by 3 so r=4

Case 2: n=22

4 + 7*22 = 3Q + r

No info about r, hence inconsistent

(A) it is!
_________________

Rgds,
TGC!
_____________________________________________________________________
I Assisted You => KUDOS Please
_____________________________________________________________________________

Intern
Joined: 24 Mar 2011
Posts: 42
Location: India
Concentration: Marketing, Operations
Schools: Schulich '16 (A)
GMAT 1: 690 Q48 V36
WE: Operations (Telecommunications)
Followers: 0

Kudos [?]: 11 [0], given: 21

Re: If n is a positive integer, and r is the remainder when [#permalink]

### Show Tags

12 Aug 2013, 11:31
As n is +ve, so to fulfill the condition that (n+1) is divisible by 3,
n can be 2, 5, 8, 11, 14
and for these values, when (4+7n) is divided by 3, it leaves remainder 0 everytime.
so (1) is sufficient.

(2) n>20 is not required as for 0<n<20, we get the same remainder as for n>20.

Correct me if this method is wrong.
Re: If n is a positive integer, and r is the remainder when   [#permalink] 12 Aug 2013, 11:31

Go to page    1   2    Next  [ 30 posts ]

Similar topics Replies Last post
Similar
Topics:
18 If n is a positive integer and r is the remainder when 5 24 Aug 2011, 00:55
10 If r is the remainder when the positive integer n is divided 8 21 Sep 2010, 21:39
5 If n is a positive integer and r is the remainder when (n - 6 26 Apr 2010, 14:39
8 If n is a positive integer and r is the remainder when (n-1)(n+1) is 8 21 Mar 2009, 10:52
19 If n is a positive integer and r is remainder, when 15 21 Apr 2008, 22:00
Display posts from previous: Sort by