283 Find the unit digit of $\left(\frac{5+\sqrt{21}}{2}\right)^{2010}$.

Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
Solve the problem, which is to find the remainder when $\left[\left(\frac{5+\sqrt{21}}{2}\right)^{2010}\right]$ is divided by 10.
Let $a_{n}=\left(\frac{5+\sqrt{21}}{2}\right)^{n}+\left(\frac{5-\sqrt{21}}{2}\right)^{n}\left(n \in \mathbf{N}_{+}\right)$.
Since $\frac{5+\sqrt{21}}{2}+\frac{5-\sqrt{21}}{2}=5$,
$$
\frac{5+\sqrt{21}}{2} \cdot \frac{5-\sqrt{21}}{2}=1 \text {, }
$$

Therefore, $\frac{5+\sqrt{21}}{2} ， \frac{5-\sqrt{21}}{2}$ are the roots of the equation $x^{2}-5 x+1=0$.
Thus, $a_{n}-5 a_{n-1}+a_{n-2}=0$, which means
$$
a_{n}=5 a_{n-1}-a_{n-2}(n \geqslant 3) \text {. }
$$

Also, $a_{1}=5, a_{2}=23$ are both integers, and from equation (1), we know that $a_{n}$ is an integer for all positive integers $n$.
Hence, $a_{2010}$ is an integer.
Since $4<\sqrt{21}<5$, then
$0<5-\sqrt{21}<1$.
Therefore, $0<\frac{5-\sqrt{21}}{2}<1$.
Thus, $0<\left(\frac{5-\sqrt{21}}{2}\right)^{2010}<1$.
Consequently, $a_{2010}$ is the smallest integer greater than $\left(\frac{5+\sqrt{21}}{2}\right)^{2010}$.
Therefore, $\left[\left(\frac{5+\sqrt{21}}{2}\right)^{2010}\right]=a_{2010}-1$.
From equation (1), we have
$$
\begin{array}{l}
a_{n} \equiv 5 a_{n-1}-a_{n-2} \equiv-a_{n-2}(\bmod 5) \\
\Rightarrow a_{n} \equiv-a_{n-2} \equiv-\left(-a_{n-4}\right) \equiv a_{n-4}(\bmod 5) \\
\Rightarrow a_{2010} \equiv a_{2} \equiv 23 \equiv 3(\bmod 5) .
\end{array}
$$

From equation (1), we also have
$$
\begin{aligned}
a_{n} & \equiv a_{n-1}+a_{n-2}(\bmod 2) \\
\Rightarrow & a_{n} \equiv a_{n-1}+a_{n-2} \\
& \equiv\left(a_{n-2}+a_{n-3}\right)+a_{n-2} \\
& \equiv 2 a_{n-2}+a_{n-3} \equiv a_{n-3}(\bmod 2) \\
\Rightarrow & a_{2010} \equiv a_{3} \equiv a_{2}+a_{1} \equiv 0(\bmod 2) .
\end{aligned}
$$

Thus, $a_{2010} \equiv 3(\bmod 5)$, and
$$
a_{2010} \equiv 0(\bmod 2) \text {. }
$$

Therefore, $a_{2010} \equiv 8(\bmod 10)$.
Hence, $\left[\left(\frac{5+\sqrt{21}}{2}\right)^{2010}\right]=a_{2010}-1$
$$
\equiv 7(\bmod 10) \text {. }
$$

Thus, 7 is the answer.