<p>The total number of divisors of 480 that are of the form \(4n + 2\), \(n \geq 0\), is equal to</p>
Step-by-Step Solution
Key Concept: Express 480 in prime factorization, then recognize that divisors of form 4n+2 are exactly those divisible by 2 but not by 4 (i.e., 2¹ × odd). Count divisors of the form 2×(odd divisor of 240).
<p><strong>Step 1:</strong> Find prime factorization of 480.</p><p>480 = 32 × 15 = 2⁵ × 3 × 5</p><p><strong>Step 2:</strong> Identify the form 4n + 2.</p><p>4n + 2 = 2(2n + 1), which means divisors must have exactly one factor of 2 (and remaining factors must be odd).</p><p><strong>Step 3:</strong> Express divisors of required form.</p><p>Any divisor of form 4n + 2 = 2¹ × d, where d is an odd divisor of (480/2) = 240 = 2⁴ × 3 × 5</p><p><strong>Step 4:</strong> Count odd divisors of 240.</p><p>Odd divisors of 240 come from 3 × 5 (ignoring all powers of 2).</p><p>Number of odd divisors = (1+1)(1+1) = 4</p><p>These are: 1, 3, 5, 15</p><p><strong>Step 5:</strong> Verify the divisors.</p><p>The divisors of 480 of form 4n + 2: 2, 6, 10, 30</p><p>Check: 2 = 4(0)+2 ✓, 6 = 4(1)+2 ✓, 10 = 4(2)+2 ✓, 30 = 4(7)+2 ✓</p><p>∴ Answer: <strong>4</strong></p>
Correct Answer: B