# What is the sum of all natural numbers ‘n’ such that 1001.

A. 3221

B. 3121

C. 3203

D. 3303

A. 3221

B. 3121

C. 3203

D. 3303

Answer

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Hint- Here, we will proceed by finding the sum of natural numbers between 100 and 200 which are divisible by either of the prime factors of the number (91) whose HCF when taken with n (any natural number) needs to be greater than 1.

Complete step-by-step answer:

HCF(91,n)>1 means that the HCF between 91 and natural number n should be greater than 1.

Let us factorize number 91 into its prime factors i.e., $91 = 13 \times 7$

As, we know that HCF between any two numbers can only be greater than 1 if there is some prime factor or product of some prime factors common between the two numbers.

For HCF(91,n)>1, the natural number n should either contain 7 as a prime factor of 13 as a prime factor or both 7 and 13 as prime factors when factorized.

Here, we have to find the sum of all the natural numbers between 100 and 200 whose HCF with 91 is greater than 1.

For this required sum we will add the sum of natural numbers between 100 and 200 which are divisible by 7 (or have 7 as a prime factor) and the sum of natural numbers between 100 and 200 which are divisible by 13 (or have 13 as a prime factor). Then, from this we will subtract the sum of natural numbers between 100 and 200 which are divisible by both 7 and 13 (or have 7 and 13 as a prime factor).

For any AP series,

nth term of an AP is given by ${a_n} = {a_1} + \left( {n - 1} \right)d{\text{ }} \to {\text{(1)}}$ where , n and d are the first term, the total number of terms and the common difference of an AP series respectively.

Sum of n terms of an AP is given by ${{\text{S}}_n} = \dfrac{n}{2}\left( {{a_1} + {a_n}} \right){\text{ }} \to {\text{(2)}}$ where is the first term of the AP and ${a_n}$ is the nth or last term of the AP.

Natural numbers between 100 and 200 which are divisible by 7 are 105,112,….,196.

Clearly, these numbers are forming an arithmetic progression with a common difference of 7.

Here ${a_1} = 105$, d=7 and ${a_n} = 196$

Using formula given by equation (1), we will get the value of n.

\[

\Rightarrow 196 = 105 + 7\left( {n - 1} \right) \\

\Rightarrow 7\left( {n - 1} \right) = 91 \\

\Rightarrow n - 1 = \dfrac{{91}}{7} = 13 \\

\Rightarrow n = 13 + 1 = 14 \\

\]

So, there are total 14 natural numbers between 100 and 200 which are divisible by 7 i.e., n=7

Using formula given by equation (2), we get

Sum of all the natural numbers between 100 and 200 which are divisible by 7 is 2107.

Natural numbers between 100 and 200 which are divisible by 13 are 104,112,….,195.

Clearly, these numbers are forming an arithmetic progression with a common difference of 13.

Here ${a_1} = 104$, d=13 and ${a_n} = 195$

Using formula given by equation (1), we will get the value of n.

\[

\Rightarrow 195 = 104 + 13\left( {n - 1} \right) \\

\Rightarrow 13\left( {n - 1} \right) = 91 \\

\Rightarrow n - 1 = \dfrac{{91}}{{13}} = 7 \\

\Rightarrow n = 7 + 1 = 8 \\

\]

So, there are total 8 natural numbers between 100 and 200 which are divisible by 13 i.e., n=8

Using formula given by equation (2), we get

Sum of all the natural numbers between 100 and 200 which are divisible by 13 is 1196.

There is only one natural number between 100 and 200 which are divisible by both 7 and 13 (i.e., natural number between 100 and 200 which is divisible by 91) is 182.

So, the final sum of all natural numbers ‘n’ such that 1001 is given by

S=2107+1196-182=3121

Hence, option B is correct.

Note- In this particular problem, there is only one natural number between 100 and 200 i.e.,182 which is divisible by both 7 and 13. This natural number 182 is counted twice when we are adding the sum of natural numbers between 100 and 200 which are divisible by 7 and the sum of natural numbers between 100 and 200 which are divisible by 13.

Complete step-by-step answer:

HCF(91,n)>1 means that the HCF between 91 and natural number n should be greater than 1.

Let us factorize number 91 into its prime factors i.e., $91 = 13 \times 7$

As, we know that HCF between any two numbers can only be greater than 1 if there is some prime factor or product of some prime factors common between the two numbers.

For HCF(91,n)>1, the natural number n should either contain 7 as a prime factor of 13 as a prime factor or both 7 and 13 as prime factors when factorized.

Here, we have to find the sum of all the natural numbers between 100 and 200 whose HCF with 91 is greater than 1.

For this required sum we will add the sum of natural numbers between 100 and 200 which are divisible by 7 (or have 7 as a prime factor) and the sum of natural numbers between 100 and 200 which are divisible by 13 (or have 13 as a prime factor). Then, from this we will subtract the sum of natural numbers between 100 and 200 which are divisible by both 7 and 13 (or have 7 and 13 as a prime factor).

For any AP series,

nth term of an AP is given by ${a_n} = {a_1} + \left( {n - 1} \right)d{\text{ }} \to {\text{(1)}}$ where , n and d are the first term, the total number of terms and the common difference of an AP series respectively.

Sum of n terms of an AP is given by ${{\text{S}}_n} = \dfrac{n}{2}\left( {{a_1} + {a_n}} \right){\text{ }} \to {\text{(2)}}$ where is the first term of the AP and ${a_n}$ is the nth or last term of the AP.

Natural numbers between 100 and 200 which are divisible by 7 are 105,112,….,196.

Clearly, these numbers are forming an arithmetic progression with a common difference of 7.

Here ${a_1} = 105$, d=7 and ${a_n} = 196$

Using formula given by equation (1), we will get the value of n.

\[

\Rightarrow 196 = 105 + 7\left( {n - 1} \right) \\

\Rightarrow 7\left( {n - 1} \right) = 91 \\

\Rightarrow n - 1 = \dfrac{{91}}{7} = 13 \\

\Rightarrow n = 13 + 1 = 14 \\

\]

So, there are total 14 natural numbers between 100 and 200 which are divisible by 7 i.e., n=7

Using formula given by equation (2), we get

Sum of all the natural numbers between 100 and 200 which are divisible by 7 is 2107.

Natural numbers between 100 and 200 which are divisible by 13 are 104,112,….,195.

Clearly, these numbers are forming an arithmetic progression with a common difference of 13.

Here ${a_1} = 104$, d=13 and ${a_n} = 195$

Using formula given by equation (1), we will get the value of n.

\[

\Rightarrow 195 = 104 + 13\left( {n - 1} \right) \\

\Rightarrow 13\left( {n - 1} \right) = 91 \\

\Rightarrow n - 1 = \dfrac{{91}}{{13}} = 7 \\

\Rightarrow n = 7 + 1 = 8 \\

\]

So, there are total 8 natural numbers between 100 and 200 which are divisible by 13 i.e., n=8

Using formula given by equation (2), we get

Sum of all the natural numbers between 100 and 200 which are divisible by 13 is 1196.

There is only one natural number between 100 and 200 which are divisible by both 7 and 13 (i.e., natural number between 100 and 200 which is divisible by 91) is 182.

So, the final sum of all natural numbers ‘n’ such that 100

S=2107+1196-182=3121

Hence, option B is correct.

Note- In this particular problem, there is only one natural number between 100 and 200 i.e.,182 which is divisible by both 7 and 13. This natural number 182 is counted twice when we are adding the sum of natural numbers between 100 and 200 which are divisible by 7 and the sum of natural numbers between 100 and 200 which are divisible by 13.

Last updated date: 01st Oct 2023

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