6. The unit digit of $\left[(1+\sqrt{2})^{10}\right]$ is $(\quad)$ (where, $[x]$ denotes the greatest integer not exceeding $x$).
(A) 1
(B) 3
(C) 5
(D) 7
6.C.

Notice that $(1+\sqrt{2})^{10}=(3+2 \sqrt{2})^{5}$.
Let $a=3+2 \sqrt{2}, b=3-2 \sqrt{2}, S_{k}=a^{k}+b^{k}$.
Since $a$ and $b$ are the roots of the equation $x^{2}-6 x+1=0$, then
$$
\begin{array}{l}
a^{n}=6 a^{n-1}-a^{n-2}, b^{n}=6 b^{n-1}-b^{n-2}, \\
S_{n}=6 S_{n-1}-S_{n-2} .
\end{array}
$$

Note that $S_{0}=2, S_{1}=6$, and use $\bar{S}$ to denote the last digit of $S$, then $\bar{S}_{n}=\overline{6 \bar{S}_{n-1}-\bar{S}_{n-2}}$.

Thus $\bar{S}_{0}=2, \bar{S}_{1}=6, \bar{S}_{2}=4, \bar{S}_{3}=8, \bar{S}_{4}=4$, $\bar{S}_{5}=6$.
Since $0<b<1$, we know $0<b^{5}<1$. Therefore, $\overline{\left[a^{5}\right]}=5$.