7. (50 points) Let the sequence $\left\{x_{n}\right\}$ satisfy
$$
x_{1}=1, x_{n+1}=4 x_{n}+\left[\sqrt{11} x_{n}\right] \text {. }
$$

Find the units digit of $x_{2012}$.
7. Clearly, $x_{2}=7$, and for any positive integer $n, x_{n}$ is a positive integer.
By the property of the floor function, we have
$$
\begin{array}{l}
4 x_{n}+\sqrt{11} x_{n}>x_{n+1}=4 x_{n}+\left[\sqrt{11} x_{n}\right] \\
>4 x_{n}+\sqrt{11} x_{n}-1 .
\end{array}
$$

Multiplying both sides of the above inequality by $4-\sqrt{11}$, we get
$$
\begin{array}{l}
5 x_{n}>(4-\sqrt{11}) x_{n+1}>5 x_{n}-(4-\sqrt{11}) \\
\Rightarrow 4 x_{n+1}-5 x_{n}<\sqrt{11} x_{n+1} \\
\quad<4 x_{n+1}-5 x_{n}+(4-\sqrt{11}) \\
\Rightarrow\left[\sqrt{11} x_{n+1}\right]=4 x_{n+1}-5 x_{n} \\
\Rightarrow x_{n+2}=4 x_{n+1}+\left[\sqrt{11} x_{n+1}\right] \\
\quad=4 x_{n+1}+\left(4 x_{n+1}-5 x_{n}\right) \\
\quad=8 x_{n+1}-5 x_{n} .
\end{array}
$$

Clearly, $x_{n+2} \equiv x_{n}(\bmod 2)$,
$$
x_{n+2} \equiv 3 x_{n+1}(\bmod 5) \text {. }
$$

By $x_{1}=1, x_{2}=7$, it is easy to prove using the second principle of mathematical induction that for any positive integer $n$,
$$
x_{n} \equiv 1(\bmod 2) \text {. }
$$

By $x_{2}=7$, it is easy to prove using the first principle of mathematical induction that for any positive integer $n \geq 2$,
$$
x_{n}=2 \times 3^{n-2}(\bmod 5) \text {. }
$$

Since $x_{2012} \equiv 1(\bmod 2)$,
$$
x_{2012} \equiv 2 \times 3^{2010} \equiv 2 \times 3^{2} \equiv 3(\bmod 5) \text {, }
$$

Therefore, the last digit of $x_{2012}$ is 3.