4
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ Hint $1$: What is the sum of the odd divisors of $n$ compared to the sum of the odd divisors of $2n$? $\endgroup$ – Michael Burr Mar 31 '18 at 4:01
  • $\begingroup$ Likely not the most elegant way, but you can find the prime factorization, 2*2*3*3*5*5*7. Now to find your odd divisors, find and multiply together the 18 combinations of odd prime factors. $\endgroup$ – Air Conditioner Mar 31 '18 at 4:08
  • $\begingroup$ @AirConditioner How is it $18$ combinations? What I did was that there are $5$ odd numbers i.e., $3, 3, 5, 5, 7$. So, $2^5$? Why is my approach wrong? $\endgroup$ – Ardent Feb 15 at 17:08
4
$\begingroup$

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

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}

$\endgroup$
1
$\begingroup$

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 $3224$.

$\endgroup$
3
  • $\begingroup$ I don't think middle school has teached those sum symbols yet. $\endgroup$ – user061703 Mar 31 '18 at 4:24
  • $\begingroup$ @TrầnThúcMinhTrí Hmm okay, I'll remove them $\endgroup$ – Jim Haddocc Mar 31 '18 at 4:24
  • $\begingroup$ No you can keep it, it may be helpful for someone else who views this page $\endgroup$ – user061703 Mar 31 '18 at 4:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.