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

Question

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

Skywalker18 · Accepted Answer

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

stonecold · Answer

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

FacelessMan · Answer

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.

manishtank1988 · Answer

My 2 cents -

If you have any question please let me know -

gracie · Answer

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

gps5441 · Answer

How did we get the last no in the series

skib · Answer

how can i solve within short time?

Posted from my mobile device

helishah142 · Answer

How did we get the last number in the series?

bushrakhan20 · Answer

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.

Prabhasam · Answer

Please change the series.. 53, 130, ... I got confused when I read it.

ScottTargetTestPrep · Answer

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

rahul16singh28 · Answer

The Terms are -

53,130, .............. So, T(n) = 77n-24........ Only D satisfies.

Vibhatu · Answer

Bunuel   ScottTargetTestPrep  Please check this approach
Two numbers are (7a+4) & (11b+9)
We can write (7a+4)= (11b+9)
solve this; a= {(11b+9)-4}/7
Will get a=7; b=4
then putting value we get 53; This is first term
Apart from that we can write=LCM(7,11)k+53 (K=1,2,3....)
below 1000; K=12; we got highest number 977.
So total number is 12+1=13
Answer:D

Kinshook · Answer

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?

n = 7k + 4 = 11m + 9; where k & m are integers
7k = 11m + 5
k = (11m+5)/7

m = 4; k = 7; n = 53
(k,m) = {(7,4), (18, 11), (29, 18), .....}
k = 7 + 11p ; where p is an integer

n<7k + 4<=1000
k <= 996/7 = 142.28

k = 7 + 11p <=142
p <= 12.27
p <= 12
p = {0,1,2,...,12}: 13 integers

We get the answer above, but k & n are listed for information below.

k = {7,18,29,40,51,62,73,84,95,106,117,128,139}
n = {53, 130, 207, 284, 361, 428, 515, 592, 669, 746, 823, 900, 977}
In the first 1000 positive integers, 12 integers exist such that they leave a remainder 4 when divided by 7, and a remainder 9 when divided by 11.

IMO D

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

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