Key Concepts and Formulas

• Prime Factorization: Expressing a number as a product of its prime factors. This is crucial for understanding the H.C.F. condition.
• Arithmetic Progression (AP) Sum Formula: The sum of an arithmetic progression is given by $S_k = \frac{k}{2}(a_1 + a_k)$, where $k$ is the number of terms, $a_1$ is the first term, and $a_k$ is the last term.
• Principle of Inclusion-Exclusion for Sums: For sets A and B, the sum of elements in $A \cup B$ is $S(A \cup B) = S(A) + S(B) - S(A \cap B)$. This principle helps avoid double-counting elements that satisfy multiple conditions.

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Step-by-Step Solution

Step 1: Understand the Condition H.C.F.(91, n) > 1

The problem requires us to find the sum of natural numbers 'n' such that $100 < n < 200$ and H.C.F.(91, n) > 1. First, we find the prime factorization of 91. $91 = 7 \times 13$. For H.C.F.(91, n) to be greater than 1, 'n' must share at least one common prime factor with 91. This means 'n' must be divisible by 7, or 'n' must be divisible by 13, or 'n' must be divisible by both 7 and 13.

Step 2: Identify the Range of Numbers

We are looking for natural numbers 'n' in the range $100 < n < 200$. This means 'n' can be any integer from 101 to 199, inclusive.

Step 3: Find the Sum of Numbers Divisible by 7 in the Given Range

Let $S_7$ be the sum of numbers 'n' such that $100 < n < 200$ and 'n' is divisible by 7. The smallest multiple of 7 greater than 100 is $7 \times 15 = 105$. The largest multiple of 7 less than 200 is $7 \times 28 = 196$. The numbers are $105, 112, \dots, 196$. This forms an arithmetic progression with first term $a_1 = 105$, common difference $d = 7$, and last term $a_k = 196$. To find the number of terms (k), we use the formula $a_k = a_1 + (k-1)d$: $196 = 105 + (k-1)7$ $91 = (k-1)7$ $13 = k-1$ $k = 14$. Now, we calculate the sum $S_7$ using the AP sum formula $S_k = \frac{k}{2}(a_1 + a_k)$: $S_7 = \frac{14}{2}(105 + 196) = 7(301) = 2107$.

Step 4: Find the Sum of Numbers Divisible by 13 in the Given Range

Let $S_{13}$ be the sum of numbers 'n' such that $100 < n < 200$ and 'n' is divisible by 13. The smallest multiple of 13 greater than 100 is $13 \times 8 = 104$. The largest multiple of 13 less than 200 is $13 \times 15 = 195$. The numbers are $104, 117, \dots, 195$. This forms an arithmetic progression with first term $a_1 = 104$, common difference $d = 13$, and last term $a_k = 195$. To find the number of terms (k): $195 = 104 + (k-1)13$ $91 = (k-1)13$ $7 = k-1$ $k = 8$. Now, we calculate the sum $S_{13}$: $S_{13} = \frac{8}{2}(104 + 195) = 4(299) = 1196$.

Step 5: Find the Sum of Numbers Divisible by Both 7 and 13 (i.e., by 91) in the Given Range

Let $S_{91}$ be the sum of numbers 'n' such that $100 < n < 200$ and 'n' is divisible by both 7 and 13, which means 'n' is divisible by their least common multiple, lcm(7, 13) = 91. The smallest multiple of 91 greater than 100 is $91 \times 2 = 182$. The largest multiple of 91 less than 200 is $91 \times 2 = 182$. The only number is 182. This forms an arithmetic progression with first term $a_1 = 182$, common difference $d = 91$, and last term $a_k = 182$. The number of terms is $k=1$. The sum $S_{91}$ is simply 182.

Step 6: Apply the Principle of Inclusion-Exclusion

We need the sum of numbers divisible by 7 OR divisible by 13. Using the Principle of Inclusion-Exclusion for sums: Sum = $S_7 + S_{13} - S_{91}$ Sum = $2107 + 1196 - 182$ Sum = $3303 - 182$ Sum = $3121$.

Common Mistakes & Tips

• Incorrect Range: Ensure you are considering numbers strictly between 100 and 200 (i.e., from 101 to 199).
• Double Counting: Without subtracting the sum of numbers divisible by both 7 and 13 (i.e., by 91), you will overcount those numbers. The Principle of Inclusion-Exclusion is essential here.
• AP Formula Application: Carefully identify the first term, last term, and the number of terms for each arithmetic progression.

Summary

The problem asks for the sum of natural numbers 'n' between 100 and 200 (exclusive) such that H.C.F.(91, n) > 1. This condition implies that 'n' must be divisible by 7 or 13. We calculated the sum of numbers divisible by 7 in the given range, the sum of numbers divisible by 13 in the given range, and the sum of numbers divisible by both 7 and 13 (i.e., by 91) in the given range. By applying the Principle of Inclusion-Exclusion, we added the sums of multiples of 7 and 13 and then subtracted the sum of multiples of 91 to avoid double-counting.

The sum of numbers divisible by 7 is 2107. The sum of numbers divisible by 13 is 1196. The sum of numbers divisible by 91 is 182. The required sum is $2107 + 1196 - 182 = 3121$.

The final answer is $\boxed{3121}$.