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Math Help - Divisors from Prime Factors

  1. #1
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    Divisors from Prime Factors

    Hello all,

    Suppose I have prime factors of a number , from the prime factors how can I find distinct divisors of a number.

    Eg :
    12 = 2*2*3 ( I have 2,2,3 as prime factors )

    I need to find divisors of 12 or atleast Number of distinct divisors of 12 ( 4 in this case )
    2,3,4,6
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  2. #2
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    Well, if n is decomposed as
    <br />
n=p1^{\alpha_1}p_2^{\alpha 2}\dots p_k^{\alpha_k}<br />
    all divisors of n are of the form
    <br />
n=p1^{\beta_1}p_2^{\beta 2}\dots p_k^{\beta_k}<br />
    where
    <br />
0\leq \beta_j\leq \alpha_j<br />
    Note that
    <br />
\beta_j has(\alpha_j+1)<br />
    possible values, hence the total number of divisors of n is
    <br />
(\alpha_1+1)(\alpha_2+2)...(\alpha_k+1)<br />
    For example 12=2^2.3^1
    <br />
\alpha_1=2,\alpha_2=1<br />
    and 12 has (2+1)(1+1)=6 divisors: 1,2,3,4,6,12.
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  3. #3
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    (a1 + 1)(a2 + 2)..... (ak + 1)
    are you sure (a2+2) or (a2+1) ?
    also assuming its all additive to 1
    say 16 = 2^4;
    whats the number of divisors from this method ?

    Tanvi

    Thanks lot for the head start !
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  4. #4
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    Of course, I meant (a2+1), it's a typo-

    For 16=2^4:

    you only have one alpha: alpha_1=4, hence (4+1) divisors: 1,2,4,8, and 16.

    Greetings
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  5. #5
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    Thanks it helped !!
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