(3) For the number $(3+\sqrt{8})^{2 n}\left(n \in \mathbf{N}^{*}\right.$ and $\left.n \geqslant 2009\right)$, let $[x]$ be the integer part of $x$. Then the remainder when $\left[(3+\sqrt{8})^{2 n}\right]$ is divided by 8 is $(\quad)$.
(A) 1
(B) 3
(C) 4
(D) 7
(3) $\mathrm{A}$ Hint: $x=(3+\sqrt{8})^{2 n}$
$$
\begin{aligned}
= & \mathrm{C}_{2 n}^{0} \cdot 3^{2 n}+\mathrm{C}_{2 n}^{1} \cdot 3^{2 n-1} \cdot(\sqrt{8}) \\
& +\mathrm{C}_{2 n}^{2} \cdot 3^{2 n-2} \cdot(\sqrt{8})^{2}+\cdots \\
= & A+\sqrt{8} B\left(A, B \in \mathbf{N}^{*}\right),
\end{aligned}
$$
$$
\begin{aligned}
y & =(3-\sqrt{8})^{2 n} \\
& =C_{2 n}^{0} \cdot 3^{2 n}-C_{2 n}^{1} \cdot 3^{2 n-1} \cdot(\sqrt{8})+C_{2 n}^{2} \cdot 3^{2 n-2} \cdot(\sqrt{8})^{2}+\cdots \\
& =A-\sqrt{8} B\left(A, B \in \mathbf{N}^{*}\right) .
\end{aligned}
$$

So $x+y=2 A$, thus $x=2 A-y=(2 A-1)+(1-y)$, where $0<1-y<1$. Therefore, $[x]=2 A-1$. Hence,
$$
\begin{aligned}
2 A-1= & x+y-1 \\
= & 2\left(\mathrm{C}_{2 n}^{0} \cdot 3^{2 n}+\mathrm{C}_{2 n}^{2} \cdot 3^{2 n-2} \cdot(\sqrt{8})^{2}\right. \\
& \left.+\mathrm{C}_{2 n}^{4} \cdot 3^{2 n-4} \cdot(\sqrt{8})^{4}+\cdots\right)-1 \\
\equiv & 2\left(3^{2 n}\right)-1 \quad(\bmod 8) \\
\equiv & 2 \cdot(8+1)^{n}-1 \quad(\bmod 8) \\
\equiv & 1(\bmod 8)
\end{aligned}
$$

Therefore, the remainder when $[x]$ is divided by 8 is 1, so the answer is A. (Question provided by Yan Hong from Xinjiang Experimental High School)