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Thread: Power of P dividing n!

  1. #1
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    Power of P dividing n!

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



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

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

    Etc....

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