Question

Sum of natural numbers between $100$ and $200$ whose HCF with $91$ should be more than $1$ is

• A. $1121$
• B. $3220$
• C. $3121$
• D. $1520$

Answer 1

The correct option is C

$3121$

Explanation for the correct option :

Step-1 : Finding the numbers between $100$ and $200$ whose $HCF$ with $91$ is more than $1$

Given numbers are : $101,102,...,199$.

We have $91=7\times 13$ i.e. $7$ and $13$ are the only improper factors of $91$. So, the numbers whose $HCF$ with $91$ is more than $1$must be either a multiple of $7$ or a multiple of $13$.

Now, the numbers between $100$ and $200$ that are the multiples of $7$ are : $105,112,...,196$.

The numbers between $100$ and $200$ that are the multiples of $13$ are : $104,117,...,195$.

The numbers between $100$ and $200$ that are the multiples of both $7$ and $13$ are basically the numbers that are multiple of $91$ are : $182$.

Step-2 : Finding the sum of the numbers between $100$ and $200$ whose $HCF$ with $91$ is more than $1$

In view of Step-1, the required sum $S$ will be

(Sum of the numbers that are multiples of $7$) $+$ (Sum of the numbers that are multiples of $13$) $-$ (Sum of the numbers that are multiples of $91$)

$=\left(105+112+\cdots 196\right)+\left(104+117+\cdots +195\right)-182$

Here, we get two $A.P.$s that are

1. $105+112+\cdots 196$ of $14$ terms with the first term $105$ and common difference $7$.
2. $104+117+\cdots +195$ of $8$ terms with the first term $104$ and common difference $13$.

Step-3 : Calculating the sums of the two obtained $A.P.$s

Formula to be used : We know that the sum of the first $n$ terms of an arithmetic series with the first term $a$ and the common difference $d$ is $\frac{n}{2}\left(2a+\left(n-1\right)d\right)$.

So, using the above formula, the sum of the first series will be

$$\begin{gathered} \frac{14}{2}\left(2\times 105+13\times 7\right) \\ =7\left(210+91\right) \\ =2107 \end{gathered}$$

and the sum of the second series will be

$$\begin{gathered} \frac{8}{2}\left(2\times 104+7\times 13\right) \\ =4\left(208+91\right) \\ =1196 \end{gathered}$$

Step-4 : Calculating the required sum

So, the required sum will be

$$\begin{gathered} S=\left(105+112+\cdots 196\right)+\left(104+117+\cdots +195\right)-182 \\ =2107+1196-182 \\ =3121 \end{gathered}$$

Hence, option (C) is the correct answer.