<p>Two finite sets have <em>m</em> and <em>n</em> \((m > n)\) elements. The number of subsets of the first set is 112 more than that of the second set. The value of <em>mn</em> is</p>
Step-by-Step Solution
Key Concept: If a set has k elements, it has 2^k subsets. Setting up the equation 2^m - 2^n = 112 and factoring as 2^n(2^(m-n) - 1) = 112 allows us to find m and n by testing divisors of 112.
<p><strong>Step 1:</strong> Write the equation for subsets. A set with k elements has 2^k subsets.</p><p>Given: 2^m - 2^n = 112, where m > n</p><p><strong>Step 2:</strong> Factor out 2^n from the left side.</p><p>2^n(2^(m-n) - 1) = 112</p><p><strong>Step 3:</strong> Find the prime factorization of 112.</p><p>112 = 16 × 7 = 2^4 × 7</p><p><strong>Step 4:</strong> Since 2^n must divide 112 and only contain powers of 2, we have 2^n ∈ {2, 4, 8, 16}.</p><p>Test n = 4: 2^4(2^(m-4) - 1) = 112</p><p>16(2^(m-4) - 1) = 112</p><p>2^(m-4) - 1 = 7</p><p>2^(m-4) = 8 = 2^3</p><p>m - 4 = 3</p><p>m = 7</p><p><strong>Step 5:</strong> Verify: 2^7 - 2^4 = 128 - 16 = 112 ✓</p><p><strong>Step 6:</strong> Calculate mn.</p><p>mn = 7 × 4 = 28</p><p>∴ Answer: B</p>
Correct Answer: B