5. Let $f(x)=a_{0}+a_{1} x+a_{2} x^{2}+a_{4} x^{3}+a_{4} x^{4}+a_{5} x^{5}$ be a polynomial in $x$ where $a_{0}, a_{1}, a_{2}, a_{3}, a_{4}$ are constants and $a_{5}=7$. When divided by $x-2004, x-2005, x-2006, x-2007$ and $x-2008$, $\mathrm{f}(x)$ leaves a remainder of $72,-30,32,-24$ and 24 respectively. Find the value of $f(2009)$.
5. Answer: 1742.
We have
$$
\begin{array}{l}
f(x)=\frac{(x-2005)(x-2006)(x-2007)(x-2008)}{(-1)(-2)(-3)(-4)}(72)+\frac{(x-2004)(x-2006)(x-2007)(x-2008)}{(1)(-1)(-2)(-3)}(-30)+ \\
\frac{(x-2004)(x-2005)(x-2007)(x-2008)}{(2)(1)(-1)(-2)}(32)+\frac{(x-2004)(x-2005)(x-2006)(x-2008)}{(3)(2)(1)(-1)}(-24)+ \\
\frac{(x-2004)(x-2005)(x-2006)(x-2007)}{(4)(3)(2)(1)}(24)+7(x-2004)(x-2005)(x-2006)(x- \\
2007)(x-2008) .
\end{array}
$$

So that $f(2009)=1742$.