The number of positive divisors of 105 is:
a. 7 b. 8 c. 9 d. 10 e. 11
Hello,
$\displaystyle 105=3*35=3*7*5$
So :
1-105
3-35
5-21
7-15
-> 8
You can also do this way :
$\displaystyle 105=3^1 5^1 7^1$
In general, the number of divisors of $\displaystyle n=\prod p_k^{\alpha_k}$ is $\displaystyle \prod (\alpha_k+1)$
---> (1+1)(1+1)(1+1)=8