<p>Find the number of odd proper divisors of the number 35700. Also, find the sum of the odd proper divisors.</p>
Step-by-Step Solution
Key Concept: Odd divisors exclude all factors of 2. Use the sum formula for divisors: multiply geometric series sums for each prime factor.
<p><strong>Step 1:</strong> Prime factorize 35700.</p><p>$35700 = 2^2 \times 3 \times 5^2 \times 7 \times 17$</p><p><strong>Step 2:</strong> Find odd proper divisors. Odd divisors must have zero factors of 2.</p><p>Number of ways to select powers of 3: $2$ ways ($3^0, 3^1$)</p><p>Number of ways to select powers of 5: $3$ ways ($5^0, 5^1, 5^2$)</p><p>Number of ways to select powers of 7: $2$ ways ($7^0, 7^1$)</p><p>Number of ways to select powers of 17: $2$ ways ($17^0, 17^1$)</p><p><strong>Step 3:</strong> Total odd divisors (including 1) = $(1 + 1)(2 + 1)(1 + 1)(1 + 1) = 24$</p><p><strong>Step 4:</strong> Odd proper divisors (excluding 1) = $24 - 1 = 23$</p><p><strong>Step 5:</strong> Sum of odd proper divisors = Sum of all odd divisors - 1</p><p>Sum of all odd divisors = $(3^0 + 3^1)(5^0 + 5^1 + 5^2)(7^0 + 7^1)(17^0 + 17^1)$</p><p>$= (4)(31)(8)(18) = 17856$</p><p>Sum of odd proper divisors = $17856 - 1 = 17855$</p>
Correct Answer: 23