How to find the sum of all the divisors of the number $38808$? I have no idea how to solve this question I have been trying to solve it for some time but I am unable to...
I tried to do this question by Brute force..but apparently there are 72 divisors of this number(well..if i could get the average then I could have multiplied it by $72$ to get the sum..but I guess that's more difficult) so writing all the divisors then adding them is surely not a good choice!!
My friend did solve it..first he wrote $38808=2^3×3^2×7^2×11$..then by using some kind of formula he wrote..
$$sum=(2^0+2^1+2^2+2^3)(3^0+3^1+3^2)(7^0+7^1+7^2)(11^0+11^1)$$ which on solving gives $$15×13×57×12=133380$$ which apparently is the right answer!!!
But he only knows the formula..(which I now know too!!) And nothing else....also the numbers above $(2^0+2^1+2^2+2^3)(3^0+3^1+3^2)(7^0+7^1+7^2)(11^0+11^1)$  are in multiplication if there was a sum then I might have figured out how to get it ...but here no clue!!!
Any suggestions/hints are appreciated...
It would be really good to have an alternate solution..but if someone explains...and how to get/prove the formula then ..that is also fine..
EDIT:most of the responses here are about those brackets trying to explain how when we will expand it we will get the sum of all divisors..Thnx..I have got that..but the question is still not finished as how can we say that only this particular order of the factors that I wrote above is going to work and all others will not!! More specifically.. If someone is writing about formula please prove it..(preferably by combinations!!)
 A: I think Khan explains the concept behind the solutiom to such problems well. Have a look at his video here.
Essentially, he explains why
$$\sum_{a = 0}^3\sum_{b=0}^2\sum_{c=0}^2\sum_{d=0}^1 2^a3^b7^c11^d = \left(\sum_{i=0}^3 2^i\right)\left(\sum_{i=0}^2 3^i\right) \left(\sum_{i=0}^2 7^i\right) \left(\sum_{i=0}^1 11^i\right)$$
A: When you distribute the multiplication over the addition, note there is a term for each choice of exponent for each prime divisor. Another way of saying that is that there is one term for each divisor of $38808$. So you are just adding up all the divisors of $38808$.
A: Represent the number as powers of prime numbers, $n=p_1^{k_1} \cdots p_r^{k_r}$. We can give the sum of divisors by a common formula which is given by
$$
( p_1^0+\cdots+p_1^{k_1}) \cdots (p_r^0+\cdots+p_r^{k_r})
$$ note $p_1,\ldots,p_r $ are primes raised to $k_1, \ldots,k_r$ respectively. Note you should also know formula for GP as terms are in GP, i.e. $a(r^n-1)/(r-1)$, where $r$ is the common ratio.
