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

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

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New post 14 Mar 2016, 09:04
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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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New post 14 Mar 2016, 11:19
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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

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

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New post 17 Mar 2016, 05:27
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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

Answer D



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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New post 11 Jun 2016, 08:46
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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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New post 30 Jul 2017, 11:02
My 2 cents -

If you have any question please let me know -
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In the first 1000 positive integers, how many integers exist such that [#permalink]

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New post Updated on: 20 Aug 2017, 21:28
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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New post 20 Aug 2017, 07:28
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.
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Re: In the first 1000 positive integers, how many integers exist such that [#permalink]

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

Posted from my mobile device
Re: In the first 1000 positive integers, how many integers exist such that   [#permalink] 27 May 2018, 04:57
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