<p>Given <span>\(100 < n < 200\)</span> and H.C.F. <span>\((91, n) > 1\)</span>. The sum of all natural numbers <span>\(n\)</span> such that <span>\(100 < n < 200\)</span> and H.C.F. <span>\((91, n) > 1\)</span> is:</p>
Step-by-Step Solution
Key Concept: Since 91 = 7 × 13, the condition H.C.F.(91, n) > 1 means n must be divisible by at least one of {7, 13}. Use inclusion-exclusion principle to count such values in the range 100 < n < 200, then sum them.
<p><strong>Step 1:</strong> Factor 91 = 7 × 13. For H.C.F.(91, n) > 1, n must be divisible by 7 or 13 (or both).</p><p><strong>Step 2:</strong> Find multiples of 7 in (100, 200): 105, 112, ..., 196. These are 7k where 15 ≤ k ≤ 28, giving 14 multiples. Sum = 7(15 + 16 + ... + 28) = 7 × (14 × 43)/2 = 7 × 301 = 2107.</p><p><strong>Step 3:</strong> Find multiples of 13 in (100, 200): 104, 117, ..., 195. These are 13m where 8 ≤ m ≤ 15, giving 8 multiples. Sum = 13(8 + 9 + ... + 15) = 13 × (8 × 23)/2 = 13 × 92 = 1196.</p><p><strong>Step 4:</strong> Find multiples of both 7 and 13 (i.e., multiples of 91) in (100, 200): only 182. Sum = 182.</p><p><strong>Step 5:</strong> By inclusion-exclusion: Total sum = 2107 + 1196 − 182 = 3121.</p><p>∴ Answer: 3121</p>
Correct Answer: 3121