Question:
What is the sum of all of the odd divisors of $6300$?

Answer:
First, we find the prime factorization of $6300$ to be $2^2 \cdot 3^2 \cdot 5^2 \cdot 7$. Note that the odd divisors of 6300 are precisely the integers of the form $3^a5^b7^c$ where $0\leq a \leq 2$, $0\leq b\leq 2$, and $0\leq c \leq 1$. Note also that distributing $(1+3+9)(1+5+25)(1+7)$ yields 18 terms, with each integer of the form $3^a5^b7^c$ (again, where $0\leq a \leq 2$, $0\leq b\leq 2$, and $0\leq c \leq 1$) appearing exactly once.  It follows that the sum of the odd divisors of 6300 is $(1+3+9)(1+5+25)(1+7)=\boxed{3224}$.