Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $p$?

share|improve this question
    
Moreover, $k \leq p \implies pk|d$. –  barak manos Aug 28 at 14:53

3 Answers 3

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,

share|improve this answer

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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.