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Thread: Smallest number with given number of factors

  1. #1
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    Find Smallest Number For Number of Factors

    I would like to know how you would fine the smallest ordinary number with exactly a certain number of factors given, for example:

    The number of factors is 6
    The smallest number with 6 factors is 12
    12 =
    • 1 * 12
    • 2 * 6
    • 3 * 4


    So the question is really:
    The number of factors is n
    The smallest number with n factors is X
    How do you find X?

    Please feel free to use an example because I think doing it all in algebra will be unreasonably difficult
    Last edited by Tzarian; Apr 26th 2012 at 11:52 AM.
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  2. #2
    Junior Member ignite's Avatar
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    Re: Smallest number with given number of factors

    Let N be any natural number and $\displaystyle N={p_1}^{a_1}{p_2}^{a_2}...{p_n}^{a_n}$ , where $\displaystyle p_1,p_2,...,p_n$ are prime numbers and $\displaystyle a_1,a_2,a_n$ are natural numbers.
    Then number of factors of N is given by $\displaystyle (a_1+1)(a_2+1)...(a_n+1)$

    You are given number of factors say k.First step would be to factorize k.That is write k as $\displaystyle (a_1+1)(a_2+1)...(a_n+1)$.
    Once you have found out $\displaystyle a_1,a_2,..,a_n$ such that $\displaystyle a_1 \ge a_2 ...\ge a_n $ , your x would be $\displaystyle 2^{a_1} 3^{a_2} 5^{a_3} .... $

    Main thing here is that you can multiple possibilities for $\displaystyle a_i 's $ and you have to optimize for that.

    Example:Given k=12,find x.
    k=3*2*2 . Corresponding x = $\displaystyle 2^{(3-1)} 3^{(2-1)} 5^{(2-1)} $=60
    k=6*2 . Corresponding x = $\displaystyle 2^{(6-1)} 3^{(2-1)} $ = 96
    k=4*3 . Corresponding x = $\displaystyle 2^{(4-1)} 3^{(3-1)} $ = 72
    So minimum x would be 60.
    Last edited by ignite; Apr 26th 2012 at 02:00 PM.
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