6. Let $a, b \in \mathbf{N}_{+}$, and $a+b \sqrt{2}=(1+\sqrt{2})^{100}$. Then the last two digits of $a b$ are $\qquad$ .
6.64.

Since $a+b \sqrt{2}=(1+\sqrt{2})^{100}$
$$
\begin{array}{l}
=(1+\sqrt{2})\left[\left((1+\sqrt{2})^{3}\right]^{33}\right. \\
=(1+\sqrt{2})(7+5 \sqrt{2})^{33},
\end{array}
$$

Therefore, by the binomial theorem we get
$$
\begin{array}{l} 
a= \sum_{i=1}^{16} \mathrm{C}_{33}^{2 i} 7^{33-2 i}(5 \sqrt{2})^{2 i}+ \\
\sqrt{2} \sum_{i=0}^{16} C_{33}^{2 i+1} 7^{33-(2 i+1)}(5 \sqrt{2})^{2 i+1}, \\
b= \frac{1}{\sqrt{2}} \sum_{i=0}^{16} C_{33}^{2 i+1} 7^{33-(2 i+1)}(5 \sqrt{2})^{2 i+1}+ \\
\sum_{i=0}^{16} C_{33}^{2 i} 7^{33-2 i}(5 \sqrt{2})^{2 i} .
\end{array}
$$

Then $a \equiv \mathrm{C}_{33}^{0} 7^{33}+\mathrm{C}_{33}^{2} 7^{31} \times 50+\sqrt{2} \mathrm{C}_{33}^{1} 7^{32} \times 5 \sqrt{2}$
$$
\begin{array}{l} 
\equiv 7^{33}+\frac{33 \times 32}{2} \times 7^{31} \times 50+33 \times 7^{32} \times 10 \\
\equiv 7 \times 49^{16}+3 \times 49^{16} \times 10 \equiv 37(50-1)^{16} \\
\equiv 37(1-16 \times 50) \equiv 37(\bmod 100), \\
b \equiv \frac{1}{\sqrt{2}}\left(C_{33}^{1} \times 7^{33} \times 5 \sqrt{2}+C_{33}^{3} \times 7^{30} \times 5^{3} \times 2 \sqrt{2}\right)+ \\
\quad C_{33}^{0} \times 7^{33}+C_{33}^{2} \times 7^{31} \times 50 \\
\equiv 33 \times 49^{16} \times 5+\frac{33 \times 32 \times 31}{3 \times 2} \times 49^{15} \times 25 \times 10+ \\
7 \times 49^{16}+\frac{33 \times 32}{2} \times 7 \times 49^{15} \times 50 \\
\equiv 65(50-1)^{16}+7(50-1)^{16} \\
\equiv 72\left(50^{2}-100+1\right)^{8} \equiv 72(\bmod 100) . \\
\text { Thus, } a b \equiv 37 \times 72 \equiv(30+7)(70+2) \\
\equiv 60+490+14 \equiv 64(\bmod 100) .
\end{array}
$$