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Math Help - factorial problem

  1. #1
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    factorial problem

    find thw largest value of n for which 125! is divisible by 6^n
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by prasum View Post
    find thw largest value of n for which 125! is divisible by 6^n
    Since there are such a lot of 2s in the prime factorisation of 125! this is equivalent to asking how many 3s are there in the prime factorisation of 125!, and you can just count them.

    CB
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    [\frac{125}{3}]+[\frac{125}{3^2}]+[\frac{125}{3^3}]+[\frac{125}{3^4}]=41+13+4=59
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  4. #4
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    Here is another way.
    Look at this site.
    Look at the exponent of 3. Why does that work?
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    Quote Originally Posted by Also sprach Zarathustra View Post
    [\frac{125}{3}]+[\frac{125}{3^2}]+[\frac{125}{3^3}]+[\frac{125}{3^4}]=41+13+4=59
    how did you get this
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  6. #6
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by prasum View Post
    how did you get this

    Legendre Formula:

    \displaystyle{ n!=\displaystyle\Pi_{p\leq n} {p^{\Sigma^{\infty}_{k=1}[\frac{n}{p^k}]} when p is prime.


    []-Floor and ceiling functions - Wikipedia, the free encyclopedia
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by prasum View Post
    how did you get this
    To expand on Also sprach Zarathustra's post in general if p is prime then the m for which p^m\| n! (this means that p^m\mid n! but p^{m+1}\nmid n!) is \displaystyle \sum_{k=1}^{\infty}\left\lfoor \frac{n}{p^k}\right\rfloor. To understand why this is try to understand why


    \displaystyle m=\left(\#\text{ of }k\leqslant n!\text{ such that }p\mid k\right)+\left(\#\text{ of }k\leqslant n!\text{ such that }p^2\mid k\right)+\left(\#\text{ of }k\leqslant n!\text{ such that }p^3\mid k\right)+\cdots


    and how this sum describes that number.
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