9・2 Given $a=3^{55}, b=4^{44}, c=5^{33}$, then we have
(A) $a<b<c$.
(B) $c<b<a$.
(C) $c<a<b$.
(D) $a<c<b$.
(China Junior High School Mathematics League, 1995)
$$
\begin{array}{lc}
\text { [Solution] } & \because a=3^{55}=\left(3^{5}\right)^{11}=243^{11}, \\
\text { and } & b=4^{44}=\left(4^{4}\right)^{11}=256^{11}, \\
\text { and } & c=5^{33}=\left(5^{3}\right)^{11}=125^{11}, \\
\therefore & c<a<b,
\end{array}
$$

Therefore, the answer is (C).
$$