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:

44% (02:39) correct 56% (02:59) wrong based on 130 sessions

HideShow timer Statistics

In the first 1000 positive integers, how many integers exist such that they leave a remainder 4 when divided by 7, and a remainder 9 when divided by 11?

Re: In the first 1000 positive integers, how many integers exist such that [#permalink]

Show Tags

14 Mar 2016, 10:19

3

This post received KUDOS

2

This post was BOOKMARKED

When integer A is divided by 7 , it leaves a remainder of 4 A= 7 q +4 A can take values 4 , 11 , 18 , 25 , 32 , 39 , 46 , 53 , 60 When integer A is divided by 11 , it leaves a remainder of 9 A= 11 p + 9 A can take values 9 , 20 , 31 , 42 , 53 , 64 The first integer which fulfills the given criteria is 53 . Similarly , we will get the next such after an interval of LCM of 7 and 11 , that is 77 The numbers are 53 , 120 .... and 977 (977= 77*12 + 53)

Number of such integers = 13

Answer D
_________________

When everything seems to be going against you, remember that the airplane takes off against the wind, not with it. - Henry Ford The Moment You Think About Giving Up, Think Of The Reason Why You Held On So Long +1 Kudos if you find this post helpful

Re: In the first 1000 positive integers, how many integers exist such that [#permalink]

Show Tags

17 Mar 2016, 04:27

1

This post was BOOKMARKED

Skywalker18 wrote:

When integer A is divided by 7 , it leaves a remainder of 4 A= 7 q +4 A can take values 4 , 11 , 18 , 25 , 32 , 39 , 46 , 53 , 60 When integer A is divided by 11 , it leaves a remainder of 9 A= 11 p + 9 A can take values 9 , 20 , 31 , 42 , 53 , 64 The first integer which fulfills the given criteria is 53 . Similarly , we will get the next such after an interval of LCM of 7 and 11 , that is 77 The numbers are 53 , 120 .... and 977 (977= 77*12 + 53)

Number of such integers = 13

Answer D

Hey i solve the question on the same approach.. But it is taking way too much time.. Any other methods?
_________________

Concentration: General Management, Entrepreneurship

GMAT 1: 510 Q22 V19

GPA: 3.41

In the first 1000 natural numbers, how many integers exist such that t [#permalink]

Show Tags

11 Jun 2016, 07:30

1

This post was BOOKMARKED

Q. In the first 1000 natural numbers, how many integers exist such that they leave a remainder 4 when divided by 7, and a remainder 9 when divided by 11?

Re: In the first 1000 natural numbers, how many integers exist such that t [#permalink]

Show Tags

11 Jun 2016, 07:46

1

This post received KUDOS

First number in this series = 53 last number in the series = 977

The series will be an arithmetic progression with first term = 53, last term = 977, and difference = 11*7 = 77 general formula for nth term => a(n) = a(1) + (n-1)d => 977 = 53 + (n-1)*77 => n-1 = 12 => n=13. Hence D.

Re: In the first 1000 positive integers, how many integers exist such that [#permalink]

Show Tags

29 Jul 2017, 06:08

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

In the first 1000 positive integers, how many integers exist such that [#permalink]

Show Tags

30 Jul 2017, 12:01

Bunuel wrote:

In the first 1000 positive integers, how many integers exist such that they leave a remainder 4 when divided by 7, and a remainder 9 when divided by 11?

A. 10 B. 11 C. 12 D. 13 E. 14

let n=dividend (n-4)/7=p (n-11)/9=q combining, 7p-11q=5 p=7 q=4 least value of n=53 let x=number of n-1 53+(7*11)x<1001 77x<948 x<12.3 x=12 12+1=13 D

Last edited by gracie on 20 Aug 2017, 20:28, edited 1 time in total.

Re: In the first 1000 positive integers, how many integers exist such that [#permalink]

Show Tags

20 Aug 2017, 06:28

How did we get the last no in the series

FacelessMan wrote:

First number in this series = 53 last number in the series = 977

The series will be an arithmetic progression with first term = 53, last term = 977, and difference = 11*7 = 77 general formula for nth term => a(n) = a(1) + (n-1)d => 977 = 53 + (n-1)*77 => n-1 = 12 => n=13. Hence D.