6. Let $a, b$ be positive integers, and $a+b \sqrt{2}$ $=(1+\sqrt{2})^{100}$. Then the units digit of $a b$ is $\qquad$
6.4.

By the binomial theorem, we have
$$
a-b \sqrt{2}=(1-\sqrt{2})^{100} \text {. }
$$

Therefore, $a=\frac{1}{2}\left((1+\sqrt{2})^{100}+(1-\sqrt{2})^{100}\right)$,
$$
\begin{array}{l}
b=\frac{1}{2 \sqrt{2}}\left((1+\sqrt{2})^{100}-(1-\sqrt{2})^{100}\right) . \\
\text { Hence } \left.a b=\frac{1}{4 \sqrt{2}}(1+\sqrt{2})^{200}-(1-\sqrt{2})^{200}\right) \\
=\frac{1}{4 \sqrt{2}}\left((3+2 \sqrt{2})^{100}-(3-2 \sqrt{2})^{100}\right) . \\
\text { Let } x_{n}=\frac{1}{4 \sqrt{2}}\left((3+2 \sqrt{2})^{n}\right. \\
\left.-(3-2 \sqrt{2})^{n}\right)(n=1,2,3,4, \cdots),
\end{array}
$$

then $x_{1}=1, x_{2}=6$.
By the identity $a^{n}-b^{n}=(a+b)\left(a^{n-1}-b^{n-1}\right)-a b$ $\left(a^{n-2}-b^{n-2}\right)$, we can obtain the recurrence relation for the sequence $\left\{x_{n}\right\}$
$$
x_{n}=6 x_{n-1}-x_{n-2} \cdot(n=3,4,5, \cdots)
$$

Thus, the last digit of $x_{n}$ is as follows:
$$
\begin{array}{l}
x_{1}=1, x_{2} \equiv 6, x_{3} \equiv 5, x_{4} \equiv 4, x_{5} \equiv 9, x_{6}=0, \\
x_{7}=1, x_{8} \equiv 6, x_{9}=5, x_{10} \equiv 4(\bmod 10) .
\end{array}
$$

From this, we can see that $x_{n+2}=x_{n+8}(\bmod 10)(n=1,2,3$, $\cdots$.

Therefore, $x_{100}=x_{6 \times 16+4} \equiv x_{4}=4(\bmod 10)$, so the last digit of $a b=x_{100}$ is 4.