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In the first 1000 positive integers, how many integers exist such that
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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
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14 Mar 2016, 11:19
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
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17 Mar 2016, 05:27
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
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11 Jun 2016, 08:46
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) + (n1)d => 977 = 53 + (n1)*77 => n1 = 12 => n=13. Hence D.



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Re: In the first 1000 positive integers, how many integers exist such that
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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
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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 (n4)/7=p (n11)/9=q combining, 7p11q=5 p=7 q=4 least value of n=53 let x=number of n1 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
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20 Aug 2017, 07:28
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) + (n1)d => 977 = 53 + (n1)*77 => n1 = 12 => n=13. Hence D.



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Re: In the first 1000 positive integers, how many integers exist such that
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27 May 2018, 04:57
how can i solve within short time?
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Re: In the first 1000 positive integers, how many integers exist such that
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31 Jul 2018, 08:23
How did we get the last number in the series?



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Re: In the first 1000 positive integers, how many integers exist such that
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31 Jul 2018, 10:09
helishah142977 is calculated by adding 53 to the last multiple of 77 that is < 1000 i.e. 53+ 77*12 = 977. Hope it helps.



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



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Re: In the first 1000 positive integers, how many integers exist such that
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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
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Re: In the first 1000 positive integers, how many integers exist such that
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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) = 77n24........ Only D satisfies.
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Re: In the first 1000 positive integers, how many integers exist such that
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