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Let $p$ be a prime number and $k$ a positive integer. Let $d$ be the smallest positive integer such that $p^k$ divides $d!$. It is true that $d$ is necessarily a multiple of $p$?

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Moreover, $k \leq p \implies pk|d$. – barak manos Aug 28 '14 at 14:53
up vote 6 down vote accepted

Suppose $d!$ is this number. Then $(d-1)!$ is not. Since $p^k$ does not divide $(d-1)!$ but it divides $d! = d \cdot (d-1)!$, then $p$ must divide $d$.

Hope that helps,

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Yes. If $p$ does not divide $d$, then $(d-1)!$ would also be divisible by $p$ since it has the same number of prime factors $p$ as $d!$ has.

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Yes. If it didn't then $d!$ would have the same power of $p$ in its factorisation as $(d-1)!$ has, thus $p^k | (d-1)!$ contradicting the minimality.

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