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Math Help - help this divisibility problem

  1. #1
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    help this divisibility problem

    The number of positive divisors of 105 is:

    a. 7 b. 8 c. 9 d. 10 e. 11
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  2. #2
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    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
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