Problem 1.3. Knowing the last digit of $2009^{2011}$ is very easy, but how many zeros precede that last digit?
Solution Problem 1.3 If $n \geq 1$,

$$
2009^{n}=(2000+9)^{n}=9^{n}+2000 k
$$

Therefore, the last three digits of $2009^{n}$ coincide with those of $9^{n}$. By the binomial expansion:

$$
\begin{aligned}
& 9^{2011}=(10-1)^{2011}=(-1)^{2011}+\left(\begin{array}{c}
2011 \\
1
\end{array}\right)(-1)^{2010} \cdot 10+ \\
& +\left(\begin{array}{c}
2011 \\
2
\end{array}\right)(-1)^{2009} \cdot 10^{2}+K \cdot 10^{3}=-1+20110-2011 \cdot 1005 \cdot 100+K \cdot 10^{3} \\
& =-202085391+K \cdot 10^{3}=609+K^{\prime} \cdot 10^{3}
\end{aligned}
$$

Thus, the answer is that 9 is the last digit and it is preceded by a single zero.