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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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.

Answer: A.

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

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.

Answer: A.

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?

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 _________________

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.

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

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 .... _________________

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

Show Tags

26 Oct 2012, 05:47

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.

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 _____________________________________________________________________________

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.

gmatclubot

Re: If n is a positive integer, and r is the remainder when
[#permalink]
12 Aug 2013, 11:31

Post your Blog on GMATClub We would like to invite all applicants who are applying to BSchools this year and are documenting their application experiences on their blogs to...

Since the value of the NZ Dollar is much lower than the Pound, foreign currency exchange rates and how to pay MBA tuition fees are obviously of much concern...