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Math Help - Proof Help

  1. #1
    kap
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    Proof Help

    Hi Guys,

    I need some help in proving this:

    The factorization of n is given as:

    <br />
n = p_{1}^ep_{2}^e...p_{k}^e<br />

    [the e's in the superscripts are e1, e2 ... ek]

    I have to prove that the number of divisors of n is equal to:

    (e_{1} + 1)(e_{2} + 1)...(e_{k} + 1)

    Need help in proving this guys.


    Thanks for your help.
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  2. #2
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    Note that any factor of n must look like p^{x_1 }  \cdot p^{x_2 }  \cdot  \cdot  \cdot p^{x_k } .
    But also note that 0\le x_j \le e_j .
    Do you see why?
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  3. #3
    kap
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    Not exactly. Some explanation please.

    Thanks for your reply.
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  4. #4
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    If 0\le n \le 3 then n=0,~1,~2,\text{ or }3that is four values for n.
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  5. #5
    kap
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    Thanks again.

    I have some little questions here:

    1)
    why is it that 1 is added to the powers of the factors?

    2)
    How do i write the proof as a whole? I mean how do i write it officially?
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  6. #6
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    Quote Originally Posted by kap View Post
    why is it that 1 is added to the powers of the factors?
    Because x_j=0,~1,~2,\cdots,~e_j that is e_j+1 values for the exponent.

    And no, I will not do this for you.
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  7. #7
    kap
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    AM not asking you to do it for me but i want you to explain to me how to do it step by step.

    thanks
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  8. #8
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    Quote Originally Posted by kap View Post
    AM not asking you to do it for me but i want you to explain to me how to do it step by step.
    That is exactly what I have done.
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