3. Find the digit in the thousands place of the number

$$
2 \cdot 7^{2009}+6 \cdot 7^{2008}+3 \cdot 7^{2007}-7^{2006}
$$
Solution. We have

$$
\begin{aligned}
n & =2 \cdot 7^{2009}+6 \cdot 7^{2008}+3 \cdot 7^{2007}-7^{2006} \\
& =7^{2006}\left(2 \cdot 7^{3}+6 \cdot 7^{2}+3 \cdot 7-1\right) \\
& =7^{2004} \cdot 7^{2} \cdot 1000 \\
& =\left(7^{4}\right)^{501} \cdot 49 \cdot 1000 \\
& =2401^{501} \cdot 49 \cdot 1000
\end{aligned}
$$

Since $2401^{501}$ has the digit 1 as its units digit, it follows that $2401^{501} \cdot 49$ ends in the digit 9, and the last four digits of $n$ are 9000. Therefore, the thousands digit of the number $n$ is 9.