<p>The sum of all natural numbers \(n\) such that \(100 < n < 200\) and H.C.F. \((91, n) > 1\) is __________.</p>
Step-by-Step Solution
Key Concept: Convert the inequality involving logarithms into a polynomial inequality by recognizing that log_n(100) < 2 is equivalent to 100 < n². This transforms the problem into finding all natural numbers n where n² > 100 and n² < 10000.
<p><strong>Step 1:</strong> Identify domain and convert first inequality.</p><p>For log_n(100) > 1 with base n > 1: log_n(100) > log_n(n) implies 100 > n, so n < 100.</p><p>For base 0 < n < 1: log_n(100) > 1 means 100 < n¹, which gives n > 100 (impossible for 0 < n < 1).</p><p>Therefore: <strong>n < 100</strong></p><p><strong>Step 2:</strong> Convert second inequality log_n(100) < 2.</p><p>For base n > 1: log_n(100) < 2 = log_n(n²) implies 100 < n², so <strong>n > 10</strong>.</p><p>For base 0 < n < 1: log_n(100) < 2 means 100 > n², so n < 10 (already excluded from Step 1).</p><p><strong>Step 3:</strong> Combine conditions.</p><p>We need: n > 1 (for logarithm base), 10 < n < 100, and n ∈ ℕ.</p><p>So: n ∈ {11, 12, 13, ..., 99}</p><p><strong>Step 4:</strong> Calculate sum.</p><p>Sum = (11 + 99) × 89 / 2 = 110 × 89 / 2 = 4895</p><p>∴ Answer: <strong>4895</strong></p>
Correct Answer: 100