<p>Sum of all even divisors of the number \(N = 2016\).</p>
Step-by-Step Solution
Key Concept: Find sum of even divisors by factoring out $2^1$ and using the divisor sum formula on remaining prime factors.
<p><strong>Solution:</strong></p><p>First, factorize $N = 2016 = 2^5 \cdot 3^2 \cdot 7^1$</p><p>Sum of all even divisors = (sum of powers of 2 from $2^1$ to $2^5$) × (sum of powers of 3 from $3^0$ to $3^2$) × (sum of powers of 7 from $7^0$ to $7^1$)</p><p>$= (2 + 2^2 + 2^3 + 2^4 + 2^5)(1 + 3 + 3^2)(1 + 7)$</p><p>$= (2 + 4 + 8 + 16 + 32)(1 + 3 + 9)(1 + 7)$</p><p>$= (62)(13)(8) = 6448$</p>
Correct Answer: B