Math Help - help this divisibility problem

1. help this divisibility problem

The number of positive divisors of 105 is:

a. 7 b. 8 c. 9 d. 10 e. 11

2. Hello,

$105=3*35=3*7*5$

So :

1-105
3-35
5-21
7-15

-> 8

You can also do this way :

$105=3^1 5^1 7^1$

In general, the number of divisors of $n=\prod p_k^{\alpha_k}$ is $\prod (\alpha_k+1)$

---> (1+1)(1+1)(1+1)=8