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

Question

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

Hello! I am a middle school student, so simply-worded answers would be super helpful.

To solve this, I tried to find the sum of the odd divisors of a few smaller numbers, like 10. I know $10 = 2 * 5$, so I thought that, for the number to be odd, I'd have to exclude 2. Therefore, including 1, the sum would be $1 + 5 = 6$. This was correct, but I think that's only because the number is so small. When I tried it again with 18, which is $3^{2} * 2$, I got a different answer from the correct sum. How should I start this problem?

Answer 1

$$6300 = 2^2\cdot3^2\cdot 5^2\cdot 7$$

The set of odd factors of $6300$ is equal to the set of factors of $3^2 \cdot 5^2 \cdot 7$.

The sum of factors of $3^2 \cdot 5^2 \cdot 7$ is

\begin{align} \sum_{a=0}^2 \sum_{b=0}^2 \sum_{c=0}^1 3^a\cdot 5^b \cdot 7^c &= \sum_{a=0}^2 3^a\sum_{b=0}^2 5^b\sum_{c=0}^1 7^c \\ &=(1+3+3^2)(1+5+5^2)(1+7) \end{align}

Answer 2

The sum of odd divisor of $6300$ is equal to the sum of all divisor of $1575$, it's highest odd factor. This is because any odd divisor of $6300$ is a divisor of $1575$, and every divisor of $1575$ is also a odd divisor of $6300$.

Now we'll want to find the sum of divisors of $a=p_1^{b_1}p_2^{b_2}..$, where I've written the prime factorization of $a$. A general factor of $a$ can be written as $p_1^{k_1}p_2^{k_2}.. $, where $0\le k_i<b_i$. Thus, sum of all divisors $a$ is $$\sum_{k_1}\sum_{k_2}...(p_1^{k_1}p_2^{k_2}....)$$ $$=\sum_{k_1}p_1^{k_1}\sum_{k_2}p_2^{k_2}....$$ $$=\frac{p_1^{b_1+1}-1}{p_1-1}\frac{p_2^{b_2+1}-1}{p_2-1}....$$

Thus, the sum of divisors of $1575=7*3^2*5^2$ is $3224$.