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

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Difficulty:   95% (hard)

Question Stats: 53% (02:58) correct 47% (03:06) wrong based on 248 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?

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

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

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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

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GRE 1: Q169 V154 Re: In the first 1000 positive integers, how many integers exist such that  [#permalink]

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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

Hey i solve the question on the same approach..
But it is taking way too much time..
Any other methods?
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Re: In the first 1000 natural numbers, how many integers exist such that t  [#permalink]

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

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My 2 cents -

If you have any question please let me know -
Attachments my 2 cents.png [ 3.39 MiB | Viewed 11402 times ]

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

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

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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  Joined: 27 May 2018
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Re: In the first 1000 positive integers, how many integers exist such that  [#permalink]

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how can i solve within short time?

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

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

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

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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

Please change the series.. 53, 130, ... I got confused when I read it.
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Re: In the first 1000 positive integers, how many integers exist such that  [#permalink]

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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.

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

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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.
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If my Post helps you in Gaining Knowledge, Help me with KUDOS.. !! Re: In the first 1000 positive integers, how many integers exist such that   [#permalink] 11 Aug 2018, 01:13
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