Computer Science, asked by swapna9393, 7 months ago

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​

Answers

Answered by deepikavim
2

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?

Answer

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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.

Answered by mahinderjeetkaur878
0

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,

=53+(n-1)*77\leq 998\\=53+77n-77\leq 998\\=77n-24\leq 998\\=77n\leq 1022\\=n\leq 1022/77\\n=13.27\\n=13

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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