from the first 1000 natural numbers how many integers exist such that it's remainder is 4 when divided by 7 and the remainder is 9 when divided by 11
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Answer:
In first 1000 natural number, how many integers exist such that they leave a remainder 4 when divided by 7, and a remainder 9 when divided by 11?
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8
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10 Answers

Naveen Shekhar, Me n Maths in a complicated relationship
Answered October 26, 2017
As per que,the required integers should be of both forms (7k+4)and (11m+9) ;k,m€N
(7k+4) series will be {11,18,25,32,39,46,53,…,998}.
It can be perceived as an AP with a=11,d=7
(11m+9) series will be {20,31,42,53,64,75,….,999}.
It can be perceived as an AP with a=20,d=11.
So,we need common terms of both the series.
# Common series would be {53,130,207,284,…}
Here,this is also an AP with d =LCM(7,11)=77 .
Last term of the common series should be less than or equal to 998 which is min(998,999).
=>53+(n-1)77 </= 998
=>53+77n-77 </= 998
=>77n-24 </= 998
=>77n < /= 1022
=>n < /= 1022/77 =13.27
=> n=13 .
Hence there are 13 such integers exist between 1 and 1000.
Answer: - Number of integers, that exist in the first 1000 natural numbers are = 13
Detailed answer: -
- Two methods are shown, method 1 is the explanation with formula and method 2 shows the solution.
Method 1:
To find the number of integers in the first 1000 natural numbers when it is divided by 7, it leaves a remainder as 4 and when divided by 11, it leaves a remainder as 9, we need to first find the first number that is common to both the conditions provided.
For the first given condition,
Numbers that leave a remainder of 4 when divided by 7 are = 4, 11, 18, 25, 32, 39, 46, 53, 60, 67, ....
For the second given condition,
Numbers that leave a remainder of 9 when divided by 11: 9, 20, 31, 42, 53, 64, .....
From the list of numbers of the two given conditions, we can see the first number which is a part of the both the conditions.
The number = 53
Then,
We need to calculate the common difference of the terms that are common to both the conditions and find number of the terms.
The common difference of the first condition = 7
And,
The difference between the terms of the second condition = 11
The numbers common to both the conditions will have a common difference, i.e., the LCM of 7 and 11.
Therefore,
The LCM of 7 and 11 = 77
Every 77th number after 53 will be a term common in both the given conditions.
So,
The terms that are common in both the given conditions can be expressed as 77k + 53.
Here,
k = the values from 0 to 12.
Therefore,
The numbers of the integers in first 1000 integers will be 13.
OR
Method 2:
Let the integers in both the cases be (7k+4) and (11m+9).
Hence,
(7k+4) series will be {11, 18, 25, 32, 39, 46, 53, …,998}.
It is an AP with a=11 and d=7
(11m+9) series will be {20, 31, 42, 53, 64, 75, ….,999}.
It is an AP with a=20 and d=11
So,
We need to find the common terms in both the cases.
Therefore,
The Common terms in both the series will be {53, 130, 207, 284, …}
Here,
It is an AP with d = the LCM of 7 and 11
Therefore,
The LCM of 7 and 11 = 77
Now,
The last term of the common numbers or terms will be less than or equal to 998, i.e., 998 and 999.
Therefore,
Therefore,
There are 13 such integers that exist between 1 to 1000 natural numbers.
To know more about the topic, visit the below links: -
https://brainly.in/question/18305982
https://brainly.in/question/3831635
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