Three, (50 points) Try to find the last two non-zero digits of 2011!.

 

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Let the last two non-zero digits of a positive integer $k$ be denoted as $f(k)$. Below, we discuss under modulo 100.
Clearly, $5 \times f(k!)$, and if $5 \times f(k), 5 \times f(l)$, then $f(k l) \equiv f(k) f(l)$. Therefore,
$$
\begin{array}{l}
f(2011!) \equiv f\left(\frac{2011!}{2000!}\right) f(2000!) \\
\equiv 68 f(2000!), \\
f(2000!) \\
\equiv f\left(\prod_{i \in A_{1}} g_{220}(i)\right) f\left(\prod_{i \in \Lambda_{2}} g_{200}(i)\right) f\left(g_{2000}(10)\right),
\end{array}
$$

where, $A_{1}=\{1,2,3,7,8,9\}, A_{2}=\{4,5,6\}$,
$$
g_{k}(i)=\prod_{j=0}^{k-1}(10 j+i) .
$$

It is easy to see that, $g_{k}(i) g_{k}(10-i) \equiv i^{k}(10-i)^{k}(i=$ $1,2,3,4)$, and when $20 \mid k$,
$$
\begin{array}{l}
g_{k}(i) \equiv i^{k} \equiv\left\{\begin{array}{ll}
1, & i=1,3,7,9 ; \\
76, & i=2,4,6,8 .
\end{array}\right. \\
\text { Then } f\left(\prod_{i \in \Lambda_{1}} g_{200}(i)\right) \equiv 76^{2} \equiv 76, \\
f\left(\prod_{i \in \lambda_{2}} g_{200}(i)\right) \\
\equiv\left(\prod_{j=0}^{10} \frac{10 j+4}{2}\right)\left(\prod_{j=0}^{100} \frac{10 j+6}{2}\right) f\left(2^{200} \prod_{j=0}^{109}(10 j+5)\right) \\
\equiv\left(\prod_{i \in A_{1} \backslash 1,91} g_{100}(i)\right) f\left(2^{200} \prod_{j=1}^{200}(2 j-1)\right) \\
\equiv 76 f\left(2^{200} \prod_{j=1}^{200}(2 j-1)\right), \\
f\left(g_{200}(10)\right)=f(200!) .
\end{array}
$$
$$
\begin{array}{l}
\quad \text { Hence } f(2000!) \\
\equiv 76 f\left(2^{200} \prod_{j=1}^{200}(2 j-1)\right) f(200!) \\
\equiv 76 f(400!) \\
\equiv 76 f\left(\prod_{i \in \Lambda_{1}} g_{40}(i)\right) f\left(\prod_{i \in \lambda_{2}} g_{40}(i)\right) f\left(g_{40}(10)\right) . \\
\text { Also } f\left(\prod_{i \in \Lambda_{1}} g_{40}(i)\right) \equiv 76, \\
f\left(\prod_{i \in \Lambda_{2}} g_{40}(i)\right) \\
\equiv\left(\prod_{i \in A_{1} \backslash 1,91} g_{20}(i)\right) f\left(2^{40} \prod_{j=1}^{40}(2 j-1)\right) \\
\equiv 76 f\left(2^{40} \prod_{j=1}^{40}(2 j-1)\right), \\
f\left(g_{40}(10)\right)=f(40!), \\
\text { Hence } f(2000!) \equiv 76 f(80!) \\
\equiv 76 f\left(\prod_{i \in \Lambda_{1}} g_{8}(i)\right) f\left(\prod_{i \in \Lambda_{2}} g_{8}(i)\right) f\left(g_{8}(10)\right) . \\
\text { And } f\left(\prod_{i \in A_{1}} g_{8}(i)\right) \equiv \prod_{i=1}^{3} g_{8}(i) g_{8}(10-i) \\
\equiv \prod_{i=1}^{3} i^{8}(10-i)^{8} \equiv 76,
\end{array}
$$
$$
\begin{array}{l}
f\left(\prod_{i \in A_{2}} g_{8}(i)\right)=\left(\prod_{\left.i \in A_{1} \backslash 1,9\right\}} g_{4}(i)\right) f\left(2^{8} \prod_{j=1}^{8}(2 j-1)\right) \\
=2^{4} \times 7^{4} \times 3^{4} \times 8^{4} f\left(2^{8} \times 15!!\right) \equiv 44, \\
f\left(g_{8}(10)\right)=f(8!)=32,
\end{array}
$$

Therefore, $f(2000!) \equiv 76^{2} \times 44 \times 32 \equiv 8$.
Thus, $f(2011!) \equiv 68 \times 8 \equiv 44$.