13.44 The smallest positive odd integer \( n \) that makes the product \( 2^{\frac{1}{7}} \cdot 2^{\frac{3}{7}} \cdots \cdots \cdot 2^{\frac{2 n+1}{7}} \) greater than 1000 is (in the product, the denominators of the exponents are all 7, and the numerators are consecutive odd integers from 1 to \( 2 n+1 \)).
(A) 7.
(B) 9.
(C) 11.
(D) 17.
(E) 19.
(27th American High School Mathematics Examination, 1976)
[Solution] Since $2^{\frac{1}{7}} \cdot 2^{\frac{3}{7}} \cdots \cdots \cdot 2^{\frac{2 n+1}{7}}=2^{\frac{(n+1)^{2}}{7}}$, and $2^{10}=1024$. We consider the value of $n$ that makes $\frac{(n+1)^{2}}{7}$ close to 10.
By $\frac{(7+1)^{2}}{7}=9 \frac{1}{7}<10<\frac{(8+1)^{2}}{7}=11 \frac{4}{7}$,
and $2^{\frac{(7+1)^{2}}{7}}=2^{9+\frac{1}{7}}<2^{9} \cdot 2^{\frac{1}{2}}<1000$,

then 8 is the smallest integer that satisfies the condition, but the problem requires an odd number,
thus $n=9$.
Therefore, the answer is $(B)$.