TITLE: Let $P(x)$ be a polynolmial with degree $2009$ and leading coefficient unity such that $P(0)=2008,P(1)=2007,P(2)=2006,\ldots,P(2008)=0$,
QUESTION [3 upvotes]: 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?

REPLY [4 votes]: 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 if $q(x) = p(x) + (x-2008)$ then $q(x) = 0$ for all integers in $[0,2008]$, which implies that $$q(x) = ax(x-1)(x-2) \ldots (x-2008) \ ,$$ whence $$p(x) = q(x) + (x-2008) = ax(x-1)(x-2) \ldots (x-2008) + (x-2008) \ .$$
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 $2008 \ldots$
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