Thread: help this divisibility problem

The number of positive divisors of 105 is:

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

Hello,



So :

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

-> 8

You can also do this way :



In general, the number of divisors of is

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