Formulae to add all the factors of a number?

If you know the number of prime factors it has, that will give you the number of overall factors won't it? (permutations of the prime factors)

I can't think how you'd extend that to adding them all up though...

$n = (a^p) \times (b^q) \times (c^r)$

number of overall factors:

$n = (p+1) \times (q+1) \times (c+1)$

Should be correct?

There is a function which does this: the divisor function. It is an "arithmetic" function which means it takes values of natural numbers to complex numbers (other natural numbers in this case)

We can investigate the properties of it.

As it turns out, the divisor function is multiplicative, which means that $\sigma (ab) = \sigma ( a)\sigma (b)$ when a and b are coprime. This means we just need to find $\sigma$ for prime values.

As it happens, when we do this, we can prove that the formula is:

$\displaystyle \boxed{ \sigma(n) = \prod_{i=1}^{r} \frac{p_{i}^{(a_{i}+1)}-1}{p_{i}-1}}$ where $\displaystyle n = \prod_{i=1}^{r}p_{i}^{a_{i}}$

$\displaystyle \prod_{i=1}^n a_i = a_1 \times a_2 \times a_3 \times \cdots \times a_n$ is just the product (over the prime divisors in this case)

Sorry for sounding kinda stupid, but can you give me an example of the formula above in practice, please? Using my common sense mostly, and currently taking only A-Level Maths, so my mathematical knowledge is very limited

Of course: Take $n = 1960 = 2^3 \cdot 5 \cdot 7^2$

Therefore $\displaystyle \sigma(1960) = \left ( \frac{2^{3+1}-1}{2-1} \right ) \left ( \frac{5^{1+1}-1}{5-1} \right ) \left ( \frac{7^{2+1}-1}{7-1} \right ) = 15 \cdot 6 \cdot 57 = \boxed{5130}$

To see where it comes from;

$\displaystyle \sigma (p^n) = 1+p+p^2+p^3+\cdots +p^n = \frac{p^{n+1}-1}{p-1}$ using the sum of a geometric progression. (When p is prime)