$13 \cdot 63$ Given the sequence $10^{\frac{1}{11}}, 10^{\frac{2}{11}}, 10^{\frac{3}{11}}, \cdots, 10^{11}$, the smallest value of $n$ for which the product of the first $n$ terms exceeds 100000 is
(A) 7 .
(B) 8 .
(C) 9 .
(D) 10 .
(E) 11 .
(16th American High School Mathematics Examination, 1965)
[Solution] Let $p=\prod_{k=1}^{n} 10^{\frac{k}{11}}=10^{s}$,
and $\quad s=\sum_{k=1}^{n} \frac{k}{11}=\frac{1}{11} \sum_{k=1}^{n} k=\frac{1}{22} n(n+1)$.
If $10^{s}>10^{5}$, then $s>5$. Or $\frac{1}{22} n(n+1)>5$,
which means $n^{2}+n-110>0$ or $(n-10)(n+11)>0$.
$$
\because n+11>0, \quad \therefore \quad n>10 \text {. }
$$

Therefore, the answer is $(E)$.