$34 \cdot 7$ For each natural number $n$, let $a_{n}=\frac{(n+9)!}{(n-1)!}$, and let $k$ be the smallest natural number such that the rightmost nonzero digit of $a_{k}$ is odd. Then the rightmost nonzero digit of $a_{k}$ is
(A) 1.
(B) 3.
(C) 5.
(D) 7.
(E) 9.
(49th American High School Mathematics Examination, 1998)
[Solution] Clearly, $a_{n}=(n+9)(n+8) \cdots(n+1) n$.
Also, the number of factor 2 in $a_{n}$ is at least $\left[\frac{10}{2}\right]+\left[\frac{10}{4}\right]+\left[\frac{10}{8}\right]=8$ times.
And $a_{n}$ has exactly 2 multiples of 5, at this time to make the rightmost non-zero digit of $a_{n}$ an odd number, $a_{n}$ must have one number at the 7th power of 5 or above.
Take $n=5^{7}-8$, so that the sum of the powers of 5 and 2 in $a_{n}$ are equal, note that
$$
5^{7}-8 \equiv 5(\bmod 16) .
$$

Now consider the rightmost non-zero digit of $a_{n}$, let $\alpha_{k}(n \leqslant k \leqslant n+9)$ be the last digit of $k$ after removing all powers of $2$ and $5$. Note that the last digit of $5^{7}$ is 5.
$\because 5^{7}-8 \equiv 5(\bmod 16)$, and $5 \times 8, \therefore 10 \times 5^{7}-8$, so $\alpha_{n}=7$.
Also, $5^{7}-7 \equiv 6(\bmod 16)$, and $5 \times 7$,
$\therefore 2 \| 5^{7}-7$ (here \| indicates $2!5^{7}-7$, but $2^{2} \times 5^{7}-7$ ),
$5 \nmid 5^{7}-8$, so $\alpha_{m+1}=\frac{25-7}{2}=9$
$$
\begin{array}{l}
\because \quad 5^{7}-6 \equiv 7(\bmod 16), \text { and } 5 \times 6, \therefore \quad 10 \nmid 5^{7}-6, \therefore \alpha_{m+2}=9, \\
\because \quad 5^{7}-5 \equiv 8(\bmod 16), \therefore 5\left\|5^{7}-5,2^{3}\right\| 5^{7}-5, \\
\therefore \quad \alpha_{n+3}=3,
\end{array}
$$

Similarly, $\alpha_{n+4}=1, \alpha_{n+5}=1, \alpha_{n+6}=3, \alpha_{n+7}=1, \alpha_{n+8}=1$, $\alpha_{n+9}=3$.
$\because 7 \times 9 \times 9 \times \cdots \times 1 \times 3$ the last digit is 9, so the first non-zero odd number on the right of $a_{k}$ is 9.
Therefore, the answer is $(E)$.