9. Suppose $P(x)$ is a monic polynomial of degree 2023 such that
$$
P(k)=k^{2023} P\left(1-\frac{1}{k}\right)
$$
for every positive integer $1 \leq k \leq 2023$. Then $P(-1)=\frac{a}{b}$, where $a$ and $b$ relatively prime integers. Compute the unique integer $0 \leq n<2027$ such that $b n-a$ is divisible by the prime 2027 .
Answer: 406
Solution: Let $n=2023$. If $P(x)=x^{n}+a_{n-1} x^{n-1}+\cdots+a_{0}$, then let
$$
R(x)=x^{n} P\left(1-\frac{1}{x}\right)=(x-1)^{n}+a_{n-1}(x-1)^{n} x+\cdots+a_{0} x^{n} .
$$

Then, note that $Q(x)=P(x)-R(x)$ is a polynomial of degree at most $n$, and it has roots $1,2, \ldots, n$, so we have $Q(x)=k(x-1) \cdots(x-n)$ for some real constant $k$. Now we determine $P(x)$ in terms of $Q(x)$. If $g(x)=1-1 / x$, then $g(g(x))=\frac{1}{1-x}$ and $g(g(g(x)))=x$. Therefore, we have
$$
\begin{aligned}
P(x)-x^{n} P\left(1-\frac{1}{x}\right) & =Q(x) \\
P\left(1-\frac{1}{x}\right)-\left(1-\frac{1}{x}\right)^{n} P\left(\frac{1}{1-x}\right) & =Q\left(1-\frac{1}{x}\right) \\
P\left(\frac{1}{1-x}\right)-\left(\frac{1}{1-x}\right)^{n} P(x) & =Q\left(\frac{1}{1-x}\right) .
\end{aligned}
$$

Adding the first equation, $x^{n}$ times the second, and $(x-1)^{n}$ times the third yields
$$
2 P(x)=Q(x)+x^{n} Q\left(\frac{x-1}{x}\right)+(x-1)^{n} Q\left(\frac{1}{1-x}\right),
$$
so
$$
\begin{aligned}
P(x)=\frac{k}{2}((x-1)(x-2) \cdots(x-n)+(0 x-1)(-1 x & -1) \cdots(-(n-1) x-1) \\
& +(-1 x+0)(-2 x+1) \cdots(-n x+(n-1))) .
\end{aligned}
$$

Therefore,
$$
P(-1)=\frac{k}{2}(-(n+1)!+0+(2 n+1)!!) .
$$

Also, since $P$ is monic, we know that
$$
1=\frac{k}{2}(1+0-n!),
$$
so
$$
P(-1)=\frac{(2 n-1)!!-(n+1)!}{1-n!} .
$$

Modulo 2027, $(n+1)!=2026!/(2026 \cdot 2025) \equiv-1 /(-1 \cdot-2) \equiv-1 / 2$ and $n!=(n+1)!/ 2024 \equiv 1 / 6$. Also, $(2 n+1)!!\equiv 0$. So our answer is
$$
\frac{1 / 2}{1-1 / 6}=\frac{3}{5} \equiv \frac{2030}{5}=406 \text {. }
$$