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Let $p$ be a prime. Show that $${n \choose p}-\bigg[\dfrac{n}{p}\bigg]$$ is divisible by $p$, for all $n\in \mathbb{N}$.

I could not attempt this problem at all. Please help. Thank you.

EDIT: Here $\bigg[ \cdot \bigg]$ is the greatest integer function.

Martin Sleziak
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Swadhin
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1 Answers1

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This is trivial if $n \leq p$, so assume $n > p$.

Note the relationship

$\binom{n}{p} \cdot p! = n \cdot (n-1) \cdots (n-p+1)$

The product on the right side is a list of p consecutive integers, so exactly one of them is divisible by $p$. In fact, that number is

$\left\lfloor \frac{n}{p} \right\rfloor p$.

The product of the remaining p-1 factors on the right side is congruent to $(p-1)!$. Thus, dividing both sides by $p$ and then reducing modulo $p$ gives

$\binom{n}{p} \cdot (p-1)! \equiv \left\lfloor \frac{n}{p} \right\rfloor (p-1)! \pmod{p}$

Now cancel the $(p-1)!$ (allowed since it is not divisible by $p$) and the result follows.

Barry Smith
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  • Why is the number on the right $\left\lfloor \frac{n}{p} \right\rfloor p$? – WLOG Jan 31 '15 at 16:53
  • It is the largest multiple of $p$ that is $\leq n$. Use the division algorithm to write $n = qp + r$ with $0 \leq r < p$. The restriction on $r$ shows that $qp$ is this multiple. On the other hand, $\frac{n}{p} = q + \frac{r}{p}$, and since $q$ is an integer and $0 \leq \frac{r}{p} < 1$, we have $\left\lfloor \frac{n}{p} \right\rfloor = q$. – Barry Smith Jan 31 '15 at 16:56
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    @WLOG See this answer for a vivid way to see it (which extends to a very simple purely arithmetical proof that binomial coefficients are integers). – Bill Dubuque Jan 31 '15 at 17:18