The sum of all natural number 'n' such that `100 lt n lt 200` and H.C.F. `(91, n) gt 1` is

Options

A. 3221
B. 3121
C. 3203
D. 3303

Answer

To solve the problem, we need to find the sum of all natural numbers \( n \) such that \( 100 < n < 200 \) and \( \text{H.C.F.}(91, n) > 1 \).

### Step-by-Step Solution:

1. **Identify the Factors of 91**:
The first step is to find the prime factorization of 91.
\[ 91 = 7 \times 13 \]
This means that \( n \) must be divisible by either 7 or 13 for the H.C.F. to be greater than 1.

2. **Find Numbers Divisible by 7**:
We need to find all numbers \( n \) in the range \( 100 < n < 200 \) that are divisible by 7.
- The smallest number greater than 100 that is divisible by 7 is:
\[ 105 = 7 \times 15 \]
- The largest number less than 200 that is divisible by 7 is:
\[ 196 = 7 \times 28 \]
- The sequence of numbers divisible by 7 between 105 and 196 is:
\[ 105, 112, 119, 126, 133, 140, 147, 154, 161, 168, 175, 182, 189, 196 \]

3. **Count the Terms Divisible by 7**:
To find the number of terms, we can use the formula for the \( n \)-th term of an arithmetic sequence:
\[ a_n = a + (n-1)d \]
where \( a = 105 \), \( d = 7 \), and \( a_n = 196 \).
\[ 196 = 105 + (n-1) \times 7 \]
\[ 196 - 105 = (n-1) \times 7 \]
\[ 91 = (n-1) \times 7 \]
\[ n-1 = \frac{91}{7} = 13 \Rightarrow n = 14 \]

4. **Calculate the Sum of Numbers Divisible by 7**:
The sum \( S \) of an arithmetic series can be calculated using the formula:
\[ S = \frac{n}{2} \times (a + l) \]
where \( n \) is the number of terms, \( a \) is the first term, and \( l \) is the last term.
\[ S = \frac{14}{2} \times (105 + 196) = 7 \times 301 = 2107 \]

5. **Find Numbers Divisible by 13**:
Now, we find all numbers \( n \) in the range \( 100 < n < 200 \) that are divisible by 13.
- The smallest number greater than 100 that is divisible by 13 is:
\[ 104 = 13 \times 8 \]
- The largest number less than 200 that is divisible by 13 is:
\[ 195 = 13 \times 15 \]
- The sequence of numbers divisible by 13 between 104 and 195 is:
\[ 104, 117, 130, 143, 156, 169, 182, 195 \]

6. **Count the Terms Divisible by 13**:
Using the same method as before:
\[ 195 = 104 + (n-1) \times 13 \]
\[ 195 - 104 = (n-1) \times 13 \]
\[ 91 = (n-1) \times 13 \]
\[ n-1 = \frac{91}{13} = 7 \Rightarrow n = 8 \]

7. **Calculate the Sum of Numbers Divisible by 13**:
\[ S = \frac{8}{2} \times (104 + 195) = 4 \times 299 = 1196 \]

8. **Find Common Terms Divisible by Both 7 and 13**:
The least common multiple of 7 and 13 is:
\[ 91 \]
The only multiple of 91 in the range \( 100 < n < 200 \) is:
\[ 182 \]

9. **Adjust the Total Sum**:
Since 182 has been counted in both sums, we need to subtract it once from the total:
\[ \text{Total Sum} = S_7 + S_{13} - \text{Common Term} \]
\[ \text{Total Sum} = 2107 + 1196 - 182 = 3121 \]

### Final Answer:
The sum of all natural numbers \( n \) such that \( 100 < n < 200 \) and \( \text{H.C.F.}(91, n) > 1 \) is \( 3121 \).