No. of factors of a number

Question

I was learning about the relationship between the prime factors of a number and I came across this way to find the number of factors of a number. It goes like this if the prime factorization of a number $x= P_1^{L_1} P_2^{L_2} P_3^{L_3} \ldots$ then its number of factors is $(L_1 + 1)(L_2 + 1)(L_3 +1) \ldots$.

My question is can someone find a proof of this?

Answer 1

The prime $P_1$ can be a factor $0$ times, or $1$ time, or $2$ times up to $L_1$ times. Thus there are $L_1 + 1$ possibilities for $P_1$.

Likewise for $P_2$.

Likewise for $P_3$.

Just multiply these cases.

Example

Let $n=20 = 2^2 \cdot 5^1$

The factors are:

$2^0 \cdot 5^0$

$2^1 \cdot 5^0$

$2^2 \cdot 5^0$

$2^0 \cdot 5^1$

$2^1 \cdot 5^1$

$2^2 \cdot 5^1$

Answer 2

This is mainly a combinatorics problem at this point as all power are integers. Now to find the number of divisors , first think how can you distinguish between two using prime factorization. Well you can give each one a code . Say you want to find the number of divisors of $7*7*3*2=X$. If a number divides $X$ it must only have the prime factors of $X$ and we can give it a code . For example $21=(0,1,1)$ is a factor here the first number of code $0$ tells us how many 2s it (it’s prime factorization) contains . Next number tells us how manu 3s its prime factorization contains . Basically we can generate a code for each divisor of a number $M = (P_1^{p_1} P_2^{p_2} P_3^{p_3}….)$ and each divisor is of form $ (P_1^{q_1} P_2^{q_2} P_3^{q_3}….)$ where $q_i$ belongs to $[0,p_i]$. And two divisors are distinct at least one digit in their code is distinct. We have $ {p_1} +1 $ ways of choosing first digit , $ {p_2} +1 $ ways of choosing second digit and so on. So in total we have $({p_1}+1) ({p_2}+1)… ({p_n}+1)$ of distinct divisors. (Maybe this is more confusing than other answer)