2. The unit digit of $\left[(\sqrt{3}+\sqrt{2})^{2006}\right]$ is $(\quad)$
A. 0
B. 1
C. 8
D. 9
2. D $(\sqrt{3}+\sqrt{2})^{2006}=(5+2 \sqrt{6})^{1003}$. Let $f(n)=(5+2 \sqrt{6})^{n}+(5-2 \sqrt{6})^{n}(n \in \mathbf{N})$, then $f(n+2)=10 f(n+1)-f(n) \equiv-f(n)(\bmod 10)$. Therefore, $f(n+4) \equiv-f(n+2) \equiv f(n)$ (mod10). So $f(n)$ is a periodic function modulo 10 with a period of 4. Thus, $f(1003) \equiv f(3)=-f(1)$ $\equiv-10 \equiv 0(\bmod 10)$