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Is it possible for a prime number $p$ to divide $i!$ or $(p-i)!$, where $i$ is an integer and $0<i<p$?

I think it's not, after experimenting with some values, however I'm not entirely certain.

Thomas Andrews
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Jason Macville
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  • Note https://en.wikipedia.org/wiki/Euclid%27s_lemma – Chris Culter Oct 10 '17 at 19:18
  • @ThomasAndrews Yes, I knew this, but thanks for the reminder. I understand the first part (i! is not divisible by p now), however, the second part (p-i)! is not divisible by p is not entirely clear for me yet. EDIT: I understand it, after reading your second comment. Many thanks! – Jason Macville Oct 10 '17 at 19:24
  • Note that you only need to show that $p$ doesn't divide $i!$, because if $0<i<p$ then $0<p-i<p$. – Thomas Andrews Oct 10 '17 at 19:24
  • It is worth mentioning that "$p$ is a prime implies that if $p\mid a\cdot b$ then $p\mid a$ or $p\mid b$" can be easily extended by induction to say "if $p\mid a_1\cdot a_2\cdot a_3\cdots a_n$ then $p\mid a_1$ or $p\mid a_2$ or $\dots$ or $p\mid a_n$" (for integers $a,b,a_1,a_2,\dots,a_n$) – JMoravitz Oct 10 '17 at 19:48

3 Answers3

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Suppose $i<p$. Any prime factor $q$ of $i!$ must be a prime factor of $j$ for some $1 \le j \le i$; this follows from the fundamental theorem of arithmetic, which says that numbers have essentially unique prime factorisations. But then $q \le j \le i < p$, so $q \ne p$. In particular, $p$ is not a prime factor of $i!$.

Likewise, if $i>0$, then $p-i<p$, and so the same is true of $(p-i)!$—just replace $i$ by $p-i$ in the above argument.

Clive Newstead
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If $p$ is prime,

$i<p \implies p \nmid i!$

$i>0 \implies p \nmid (p-i)!$

Matt
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If $p\mid ab$ then $p\mid a$ or $p\mid b$.

In comments, you say you've seen how $p$ is not a divisor of $i!$ if $0<i<p$.

But if $0<i<p$ then $0<p-i<p$, so it is also not a divisor of $(p-i)!$ by the same reasoning.

Thomas Andrews
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