$34 \cdot 4$ For positive integers $n$ and $a$, define $n_{a}!$ as
$$
n_{a}!=n(n-a)(n-2 a)(n-3 a) \cdots(n-k a),
$$

where $k$ is the largest integer for which $n>k a$. Then the quotient of $72_{8}! \div 182!$ is
(A) $4^{5}$.
(B) $4^{6}$.
(C) $4^{8}$.
(D) $4^{9}$.
(E) $4^{12}$.
(24th American High School Mathematics Examination, 1973)
[Solution] $72_{8}!=72 \cdot 64 \cdot 56 \cdot 48 \cdot 40 \cdot 32 \cdot 24 \cdot 16 \cdot 8$
$$
=8^{9} \cdot 9!,
$$

and
$$
\begin{aligned}
182! & =18 \cdot 16 \cdot 14 \cdot 12 \cdot 10 \cdot 8 \cdot 6 \cdot 4 \cdot 2 \\
& =2^{9} \cdot 9!,
\end{aligned}
$$

then $72_{8}! \div 18_{2}!=8^{9} \div 2^{9}=4^{9}$.
Therefore, the answer is $(D)$.