<p>Find the number of odd proper divisors of \(3^p \times 6^m \times 21^n\).</p>
Step-by-Step Solution
Key Concept: First, express the number in prime factorization form: 3^p × 6^m × 21^n = 3^(p+m+n) × 2^m × 7^n. Odd divisors exclude all factors of 2, so count divisors of 3^(p+m+n) × 7^n only. Proper divisors exclude the number itself.
<p><strong>Step 1:</strong> Express in prime factorization:<br>3^p × 6^m × 21^n = 3^p × (2·3)^m × (3·7)^n = 2^m × 3^(p+m+n) × 7^n</p><p><strong>Step 2:</strong> For odd divisors, exclude the factor 2^m. Odd divisors have form 3^a × 7^b where:<br>0 ≤ a ≤ (p+m+n) and 0 ≤ b ≤ n</p><p><strong>Step 3:</strong> Total divisors (odd) = (p+m+n+1)(n+1)<br>This counts all combinations including 3^0 × 7^0 = 1 and 3^(p+m+n) × 7^n (the number itself)</p><p><strong>Step 4:</strong> Proper divisors exclude the number itself, so subtract 1:<br>Number of odd proper divisors = (p+m+n+1)(n+1) - 1</p>
Correct Answer: (p + m + n + 1)(n + 1) - 1