The number of positive divisors of 105 is:

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

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- Apr 29th 2008, 05:08 AMsri340help this divisibility problemThe number of positive divisors of 105 is:

a. 7 b. 8 c. 9 d. 10 e. 11 - Apr 29th 2008, 05:13 AMMoo
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