6. The integer $M=2003^{2004}{ }^{2005} \underbrace{2008}$ when divided by 80 leaves a remainder of ( ).
(A) 1
(B) 11
(C) 31
(D) 79
6.A.

Let $m=2004^{2000^{2008}}, p=2005^{2000^{2002^{2008}}}$.
Then $m=2004^{p}=4^{p} \times 501^{p}=4 n, n=4^{p-1} \times 501^{p}$.
Thus, by the binomial theorem we have
$$
\begin{array}{l}
M=2003^{m}=(2000+3)^{m} \\
=2000^{m}+\mathrm{C}_{m}^{1} \cdot 2000^{m-1} \cdot 3+\cdots+ \\
\mathrm{C}_{m}^{m-1} \cdot 2000 \cdot 3^{m-1}+3^{m} .
\end{array}
$$

Therefore, the remainder when $M$ is divided by 80 is the same as the remainder when $3^{m}$ is divided by 80.
$$
\begin{array}{l}
\text { Also, } 3^{m}=3^{4 n}=81^{n}=(80+1)^{n} \\
=80^{n}+C_{n}^{1} \cdot 80^{n-1}+\cdots+C_{n}^{n-1} \cdot 80+1,
\end{array}
$$

Hence, the remainder when $M$ is divided by 80 is 1.