6. Let $f(x)$ represent a fourth-degree polynomial in $x$. If
$$
f(1)=f(2)=f(3)=0, f(4)=6, f(5)=72 \text {, }
$$

then the last digit of $f(2010)$ is $\qquad$
6. 2 .

Since $f(1)=f(2)=f(3)=0$, the quartic polynomial can be set as
$$
\begin{array}{l}
f(x)=(x-1)(x-2)(x-3)(a x+b) . \\
\text { By } f(4)=6, f(5)=72, \text { we get } \\
\left\{\begin{array} { l } 
{ 6 ( 4 a + b ) = 6 , } \\
{ 2 4 ( 5 a + b ) = 7 2 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a=2, \\
b=-7 .
\end{array}\right.\right.
\end{array}
$$

Then $f(2010)$
$$
\begin{array}{l}
=(2010-1)(2010-2)(2010-3)(2010 \times 2-7) \\
=2009 \times 2008 \times 2007 \times 4013 .
\end{array}
$$

Therefore, the last digit of $f(2010)$ is 2.