$10 \cdot 67$ Given the sequence $10^{\frac{1}{11}}, 10^{\frac{2}{11}}, 10^{\frac{3}{11}}, 10^{\frac{4}{11}}, \cdots, 10^{\frac{n}{11}}$. The positive integer $n$ that makes the product of the first $n$ terms of the sequence equal to 1000000 is
(A) does not exist.
(B) 8 .
(C) 9 .
(D) 10 .
(E) 11 .
(22nd American High School Mathematics Examination, 1971)
[Solution] Since $1000000=10^{6}$, the sum of the exponents of the first $n$ terms of the sequence should be 6.
That is, $\frac{n(n+1)}{22}=6$, or $n(n+1)=132$,
$$
\therefore \quad n=11 \text {. }
$$

Therefore, the answer is $(E)$.