Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st.

It is currently 20 Jul 2019, 00:35

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

In the first 1000 positive integers, how many integers exist such that

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 56300
In the first 1000 positive integers, how many integers exist such that  [#permalink]

Show Tags

New post 14 Mar 2016, 09:04
2
23
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

53% (02:58) correct 47% (03:06) wrong based on 248 sessions

HideShow timer Statistics


Most Helpful Community Reply
Verbal Forum Moderator
User avatar
V
Status: Greatness begins beyond your comfort zone
Joined: 08 Dec 2013
Posts: 2356
Location: India
Concentration: General Management, Strategy
Schools: Kelley '20, ISB '19
GPA: 3.2
WE: Information Technology (Consulting)
GMAT ToolKit User Reviews Badge CAT Tests
Re: In the first 1000 positive integers, how many integers exist such that  [#permalink]

Show Tags

New post 14 Mar 2016, 11:19
3
3
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
General Discussion
Current Student
User avatar
D
Joined: 12 Aug 2015
Posts: 2609
Schools: Boston U '20 (M)
GRE 1: Q169 V154
GMAT ToolKit User
Re: In the first 1000 positive integers, how many integers exist such that  [#permalink]

Show Tags

New post 17 Mar 2016, 05:27
1
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?
_________________
Manager
Manager
User avatar
Joined: 19 Dec 2015
Posts: 110
Location: United States
GMAT 1: 720 Q50 V38
GPA: 3.8
WE: Information Technology (Computer Software)
Re: In the first 1000 natural numbers, how many integers exist such that t  [#permalink]

Show Tags

New post 11 Jun 2016, 08:46
1
1
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.
Manager
Manager
avatar
G
Joined: 14 Oct 2012
Posts: 160
Reviews Badge
Re: In the first 1000 positive integers, how many integers exist such that  [#permalink]

Show Tags

New post 30 Jul 2017, 11:02
My 2 cents -

If you have any question please let me know -
Attachments

my 2 cents.png
my 2 cents.png [ 3.39 MiB | Viewed 11402 times ]

VP
VP
avatar
P
Joined: 07 Dec 2014
Posts: 1207
In the first 1000 positive integers, how many integers exist such that  [#permalink]

Show Tags

New post Updated on: 20 Aug 2017, 21:28
1
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

Originally posted by gracie on 30 Jul 2017, 13:01.
Last edited by gracie on 20 Aug 2017, 21:28, edited 1 time in total.
Manager
Manager
avatar
B
Joined: 04 May 2014
Posts: 158
Location: India
WE: Sales (Mutual Funds and Brokerage)
Re: In the first 1000 positive integers, how many integers exist such that  [#permalink]

Show Tags

New post 20 Aug 2017, 07:28
1
1
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.
Intern
Intern
avatar
Joined: 27 May 2018
Posts: 1
Re: In the first 1000 positive integers, how many integers exist such that  [#permalink]

Show Tags

New post 27 May 2018, 04:57
how can i solve within short time?

Posted from my mobile device
Intern
Intern
avatar
B
Joined: 27 Dec 2016
Posts: 5
Re: In the first 1000 positive integers, how many integers exist such that  [#permalink]

Show Tags

New post 31 Jul 2018, 08:23
How did we get the last number in the series?
Intern
Intern
avatar
B
Joined: 27 Aug 2014
Posts: 3
Re: In the first 1000 positive integers, how many integers exist such that  [#permalink]

Show Tags

New post 31 Jul 2018, 10:09
1
helishah142

977 is calculated by adding 53 to the last multiple of 77 that is < 1000 i.e. 53+ 77*12 = 977.

Hope it helps.
Intern
Intern
avatar
B
Joined: 04 Apr 2018
Posts: 1
Re: In the first 1000 positive integers, how many integers exist such that  [#permalink]

Show Tags

New post 31 Jul 2018, 10:51
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


Please change the series.. 53, 130, ... I got confused when I read it.
Target Test Prep Representative
User avatar
D
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 6968
Location: United States (CA)
Re: In the first 1000 positive integers, how many integers exist such that  [#permalink]

Show Tags

New post 10 Aug 2018, 19:07
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


We need to find the smallest integer that satisfies both conditions:

Numbers that leave a remainder of 4 when divided by 7 are:

4, 11, 18, 25, 32, 39, 46, 53, ...

Numbers that leave a remainder of 9 when divided by 11 are:

20, 31, 42, 53, …

We see that 53 is the first number that satisfies both conditions. To find the subsequent numbers that also satisfy both conditions we keep adding the LCM of 7 and 11, i.e., 77, to (and beginning with) 53. So the numbers, including 53, are:

53, 130, 207, 284, 361, 438, 515, 592, 669, 746, 823, 900, and 977

So there are a total of 13 such numbers.

Answer: D
_________________

Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com
TTP - Target Test Prep Logo
122 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Director
Director
avatar
P
Joined: 31 Jul 2017
Posts: 515
Location: Malaysia
Schools: INSEAD Jan '19
GMAT 1: 700 Q50 V33
GPA: 3.95
WE: Consulting (Energy and Utilities)
Re: In the first 1000 positive integers, how many integers exist such that  [#permalink]

Show Tags

New post 11 Aug 2018, 01:13
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


The Terms are -

53,130, .............. So, T(n) = 77n-24........ Only D satisfies.
_________________
If my Post helps you in Gaining Knowledge, Help me with KUDOS.. !!
GMAT Club Bot
Re: In the first 1000 positive integers, how many integers exist such that   [#permalink] 11 Aug 2018, 01:13
Display posts from previous: Sort by

In the first 1000 positive integers, how many integers exist such that

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne