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Math Help - Power of P dividing n!

  1. #1
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    Power of P dividing n!

    If x>0 is a real number define [x] to be greatest integer less than or equal to x. If p is a prime, show that the power of p which exactly divides n! is given by \Bigl[\frac{n}{p}\Bigr] + \Bigl[\frac{n}{p^2}\Bigr]+ \cdots + \Bigl[\frac{n}{p^k}\Bigr]+ \cdots
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  2. #2
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    Quote Originally Posted by Chandru1 View Post
    If x>0 is a real number define [x] to be greatest integer less than or equal to x. If p is a prime, show that the power of p which exactly divides n! is given by \Bigl[\frac{n}{p}\Bigr] + \Bigl[\frac{n}{p^2}\Bigr]+ \cdots + \Bigl[\frac{n}{p^k}\Bigr]+ \cdots



    Suppose kp\le n<(k+1)p\,,\,k\in\mathbb{N} \Longrightarrow k\le \frac{n}{p}<\frac{k+1}{k}p\Longrightarrow \left[\frac{n}{p}\right]=k= the number of times p enters ONCE in n\Longrightarrow k= the number of times p divides n!

    Suppose mp^2\le n<(m+1)p^2\,,\,m\in\mathbb{N} \Longrightarrow m\le \frac{n}{p^2}<\frac{m+1}{m}p^2\Longrightarrow \left[\frac{n}{p^2}\right]=m= the number of times p^2 enters TWICE in n\Longrightarrow m= the number of times p^2 divides n!

    Etc....

    Tonio
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