1. ## sum of factors

What is the sum of all factors of 45000 which are divisible by 10

2. That would be 10 times the sum of all the factors of 4500. The sum-of-factors function $\displaystyle \sigma$ has the following property: if $\displaystyle n=p_1^{k_1}\cdots p_r^{k_r}$ where the $\displaystyle p_i$ are distinct primes, then

$\displaystyle \sigma(n)\ =\ \prod_{i\,=\,1}^r\frac{p_i^{k_i+1}-1}{p_i-1}$

Thus $\displaystyle \sigma(4500)\,=\,\sigma(2^23^25^3)\,=\,\left(\frac {2^3-1}{2-1}\right)\left(\frac{3^3-1}{3-1}\right)\left(\frac{5^4-1}{5-1}\right)\,=\,7\cdot13\cdot156\,=\,14196.$ So the answer to the problem is 141960.

3. can u give an easy method

4. I don’t see anything wrong with the proof, particularly the formula

$\displaystyle \sigma(n)\ =\ \prod_{i\,=\,1}^r\frac{p_i^{k_i+1}-1}{p_i-1}$

Shall we prove this formula? Then perhaps you’ll feel less afraid of using it.

First, we prove that if $\displaystyle \gcd(m,n)=1$ then $\displaystyle \sigma(mn)=\sigma(m)\sigma(n).$ Note that a divisor $\displaystyle d$ of $\displaystyle mn$ can be written as $\displaystyle d=ef$ where $\displaystyle e\mid m$ and $\displaystyle f\mid n$ and $\displaystyle \gcd(e,f)=1.$ Hence

$\displaystyle \sigma(mn)\,=\,\sum_{d\,\mid\,mn}d$

$\displaystyle =\,\sum_{e\,\mid\,m\,f\,\mid\,n}ef$

$\displaystyle =\,\sum_{e\,\mid\,m}e\,\sum_{f\,\mid\,n}f$

$\displaystyle =\,\sigma(m)\sigma(n)$

Next, not that if $\displaystyle p$ is a prime and $\displaystyle k$ a positive integer, $\displaystyle \sigma(p^k)=1+p+p^2+\cdots+p^k=\frac{p^{k+1}-1}{p-1}.$

Combining the two results established above gives the required formula for the sum-of-factors function $\displaystyle \sigma.$