BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach Starts Oct 17
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Chris Peckover, 100th-Percentile GMAT Scorer

Oct 17 · Chris Peckover
Sat · 11:00 AM to 2:00 PM ET
Oct 20 · Chris Peckover
Tue, Thu · 8:00 to 10:00 PM ET
Oct 25 · Josh Braslow
Sun · 1:00 to 4:00 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemand 5 seats left Start anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions
Start free 5-day trial
Read 559 reviewsTarget Test Prep

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

Remainder R

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Data Sufficiency |
User avatar
Master | Next Rank: 500 Posts
Posts: 472
Joined: Sun Mar 29, 2009 6:54 pm
Thanked: 56 times

by ssmiles08 » Fri Aug 14, 2009 7:04 am
I have issues with remainders as well. I got A for this problem.

the remainder equation with question stem plugged in:

4 + 7N = 3Q + R

here we know that R can have at most 3 values: 0, 1, 2.

1) n + 1 = 3P + 0

N = 3P - 1

substitute N into the first equation: 4 + 7(3P - 1) = 3Q + R

21P - 7 + 4 = 3Q + R
21P - 3 -3Q = R
3(7P - 1 - Q) = R

Now we know all variables are integers; which means R is an integer as well.

divide both sides by 3, and we find out that R is divisble by 3. The only value R can be is 0. SUFFICIENT.

2) N > 20. doesnt limit R. R can still be 0, 1, 2 depending on N.

IMO (A).
You got a dream... You gotta protect it. People can't do somethin' themselves, they wanna tell you you can't do it. If you want somethin', go get it. Period.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 208
Joined: Sat Jan 31, 2009 11:32 am
Location: Mumbai
Thanked: 2 times

by vikram_k51 » Fri Aug 14, 2009 9:19 am
If n is a positive integer and r is the remainder when 4+7n divided by 3, what is the value of r?

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


Will be A.

If n+1 is divisible by 3 then n will always give a rem of 2.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 12
Joined: Fri Jul 24, 2009 7:43 am
Location: China Beijing
GMAT Score:480

by robbie523 » Fri Aug 14, 2009 9:42 am
well, for this question, there is not many short cuts, since 4+7n is not a number can be easily broken down... to see if it fit.

well, stm one, use number like 2, 5, 8, 11, to plug in and check the result, and plug 4 in is sure enough to prove something yes or no
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 472
Joined: Sun Mar 29, 2009 6:54 pm
Thanked: 56 times

by ssmiles08 » Fri Aug 14, 2009 10:01 am
vikram_k51 wrote:
Will be A.

If n+1 is divisible by 3 then n will always give a rem of 2.

How are you getting a remainder of 2??

anyway i guess you can plug in numbers as well and don't really have to solve the equation completely to find out the answer is A.
You got a dream... You gotta protect it. People can't do somethin' themselves, they wanna tell you you can't do it. If you want somethin', go get it. Period.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 208
Joined: Sat Jan 31, 2009 11:32 am
Location: Mumbai
Thanked: 2 times

by vikram_k51 » Fri Aug 14, 2009 10:42 am
ssmiles08 wrote:
vikram_k51 wrote:
Will be A.

If n+1 is divisible by 3 then n will always give a rem of 2.

How are you getting a remainder of 2??

anyway i guess you can plug in numbers as well and don't really have to solve the equation completely to find out the answer is A.
n=5,8,11,14...all give aremainder of 2 when divided by 3
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 472
Joined: Sun Mar 29, 2009 6:54 pm
Thanked: 56 times

by ssmiles08 » Fri Aug 14, 2009 10:59 am
vikram_k51 wrote:
ssmiles08 wrote:
vikram_k51 wrote:
Will be A.

If n+1 is divisible by 3 then n will always give a rem of 2.

How are you getting a remainder of 2??

anyway i guess you can plug in numbers as well and don't really have to solve the equation completely to find out the answer is A.
n=5,8,11,14...all give aremainder of 2 when divided by 3
4 + 7N

4 + 5*7 = 39.
4 + 8*7 = 60
4 + 11*7 = 81
4 + 14*7 = 102

39, 60, 81 and 102 are all divisible by 3. which give the remainder of 0.

let me know if you think otherwise...and please show your work.
You got a dream... You gotta protect it. People can't do somethin' themselves, they wanna tell you you can't do it. If you want somethin', go get it. Period.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 1035
Joined: Wed Aug 27, 2008 10:56 pmFollowed by: 1 member
Thanked: 104 times

by scoobydooby » Fri Aug 14, 2009 12:01 pm
stment 1. n+1 is divisible by 3
=>n+1=3k
=>n=3k-1.
=>n=2, 5, 8,.... (as n>0)


so 4+7n =18, 39, 60... divisible by 3, so leaves no remainder
sufficient


stment 2. n>20
remainder can still be 0, 1, 2 depending on n.

hence, A
Last edited by scoobydooby on Fri Aug 14, 2009 12:49 pm, edited 1 time in total.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 472
Joined: Sun Mar 29, 2009 6:54 pm
Thanked: 56 times

by ssmiles08 » Fri Aug 14, 2009 12:15 pm
scoobydooby wrote:stment 1. n+1 is divisible by 3
=>n+1=3k
=>n=3k+1.
=>n=1, 4, 7, 10.... (as n>0)
if n+1 = 3k

shouldn't n = 3k - 1??? and not 3k + 1
You got a dream... You gotta protect it. People can't do somethin' themselves, they wanna tell you you can't do it. If you want somethin', go get it. Period.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 1035
Joined: Wed Aug 27, 2008 10:56 pmFollowed by: 1 member
Thanked: 104 times

by scoobydooby » Fri Aug 14, 2009 12:42 pm
oops. thanks. will edit :)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 752
Joined: Sun May 17, 2009 11:04 pm
Location: Tokyo
Thanked: 81 times
GMAT Score:680

by tohellandback » Fri Aug 14, 2009 9:57 pm
IMO A
1)meean n is of the form 3k+2

3+1 +7(3k+2)
=3+21k+15
all terms are divisible by 3. remainder 0

2)it depends upon the number n, so not sufficient
Last edited by tohellandback on Sun Aug 16, 2009 8:29 pm, edited 1 time in total.
The powers of two are bloody impolite!!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 65
Joined: Sun Oct 19, 2008 2:18 pm
Thanked: 1 times

Re: Remainder R

by yezz » Sat Aug 15, 2009 1:55 pm
joyseychow wrote:If n is a positive integer and r is the remainder when 4+7n divided by 3, what is the value of r?

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


4+7n = (3+6n)+(n+1)

from 1

n+1 is dev by 3 thus as 3+6n is devis by 3 the whole expression 4+7n is too
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 43
Joined: Thu Mar 12, 2009 8:56 pmFollowed by: 1 member
Thanked: 10 times
GMAT Score:730

by jjk » Mon Aug 17, 2009 4:14 pm
vikram_k51 wrote:If n is a positive integer and r is the remainder when 4+7n divided by 3, what is the value of r?

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


Will be A.

If n+1 is divisible by 3 then n will always give a rem of 2.
The problem does not ask you the value of r when n is divided by 3. It asks you the value of r when 4 + 7n is divided by 3.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Newbie | Next Rank: 10 Posts
Posts: 2
Joined: Mon Oct 03, 2011 12:25 pm

by rockfeld » Wed May 09, 2012 1:49 am
I think the easiest approach would be:
1. 4+7n=4+4n+3n=4*(1+n)+3n
2. (1) 1+n is a multiple of 3 => if we divide 4*(1+n)+3n by 3 then the remainder would be 1/3. That's it. Suff.
3. (2) not suff info.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
GMAT Instructor
Posts: 3225
Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members
Location: Toronto
Thanked: 1710 times
GMAT Score:800

by Stuart@KaplanGMAT » Wed May 09, 2012 4:44 am
joyseychow wrote:If n is a positive integer and r is the remainder when 4+7n divided by 3, what is the value of r?

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


Help! I always have difficulty with remainders.
Picking numbers is a very efficient way to solve number property DS questions.

From the question stem, we know that (4+7n)/3 has a remainder of r; we want to know that value. What do we need? Information about n.

1) n+1 is a multiple of 3.

If n=2 (we're allowed to pick 2 since 2+1 is a multiple of 3), then (4+14)/3 = 18/3 = 6rem0
If n=5 (we're allowed to pick 5 since 5+1 is a multiple of 3), then (4+35)/3 = 39/3 = 13rem0

at this point you might already be conviced that you'll always get the same answer, but we could try one more just to be safe:

If n=8 (we're allowed to pick 8 since 8+1 is a multiple of 3), then (4+56) = 60/3 = 20rem0

For all 3 plug-ins we get r=0.. sufficient!

2) n > 20

If n=21, then (4+147)/3 = 151/3 = 50rem1

Note: if you really understand DS, you can actually stop right now. Here's a fundamental rule to remember:

If the answer is (D), then both statements will always give the same answer to the question.

Since we got a remainder of 1, and since statement (1) guaranteed a remainder of 0, we know that (2) cannot be sufficient!

However, if you don't see that logic, then you just pick another number:

If n=22, then (4+154)/3 = 158/3 = 52rem2.

(2) has now given us r=1 AND r=2; more than one possible value, therefore insufficient.

(1) is suff, (2) is infuff: choose A!
Image

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1