# Let $P(x)$ be a polynolmial with degree $2009$ and leading coefficient unity such that $P(0)=2008,P(1)=2007,P(2)=2006,\ldots,P(2008)=0$,

Let $P(x)$ be a polynolmial with degree $2009$ and leading coefficient unity such that $P(0)=2008,P(1)=2007,P(2)=2006,\ldots,P(2008)=0$,then the value of $P(2009)=n!-a$ where $n!$ is $n$ factorial,$n,a$ are natural numbers.Find $n+a.$

If i let $P(x)=a_{2009}x^{2009}+a_{2008}x^{2008}+a_{2007}x^{2007}+\cdots+a_1 x+a_0$ with $a_{2009}=1$,then it is very difficult to find $P(2009)$ with the given data.What should i do?

#### Solutions Collecting From Web of "Let $P(x)$ be a polynolmial with degree $2009$ and leading coefficient unity such that $P(0)=2008,P(1)=2007,P(2)=2006,\ldots,P(2008)=0$,"

You want a way to find $P(2009)$ in terms of a factorial, but it is an extra hint! The question tells you that $P(2009)$ can be viewed as a sum of a factorial and another number! This suggest the polynomial should look something like $(x-1)(x-2) \ldots$ now let’s give up heuristic reasoning and focus on algebra $\ldots$

Note that
\begin{align}
\forall x \in [0,2008] \quad &q(x) = p(x) – (x-2008) = 0\\
&\Rightarrow q(x) = ax(x-1)(x-2) \ldots (x-2008)\\
&\Rightarrow p(x) = q(x) + (x-2008)\\
&\Rightarrow p(x) = ax(x-1)(x-2) \ldots (x-2008) + (x-2008)
\end{align}
but we don’t know $a \ldots$ wait wait, leading coefficient unity tells us that $a = 1$! Hence we got a solution
$$p(2009) = a\cdot 2009 \cdot (2009-1) \cdot \ldots \cdot 1 + (2009 – 2008) = 1 \cdot 2009! + 1$$
thus
$$n = 2009, a = -1\\ \Rightarrow n+a = 2008$$
Somehow I think that’s from a mathematical contest in $2008 \ldots$