(9) Let $\left\{a_{n}\right\}$ be a geometric sequence with common ratio $q \in(1,2)$. When $1 \leqslant n \leqslant 6$, $a_{n}$ is a positive integer. Then the minimum value of $a_{6}$ is $\qquad$ .
9243 Hint: $a_{n}=a_{1} q^{n-1}$, since $a_{1}, \cdots, a_{6}$ are all positive integers, $q$ must be a rational number. Let $q=\frac{s}{r}$, where $s, r$ are positive integers, and $(s, r)=1, 2 \leqslant r<s<2 r$, then $a_{6}=a_{1} \frac{s^{5}}{r^{5}}$. From $(s, r)=1$ we get $\left(s^{5}, r^{5}\right)=1$, so $a_{1}$ is an integer multiple of $r^{5}$. Therefore, $a_{6}=a_{1} \frac{s^{5}}{r^{5}} \geqslant s^{5} \geqslant 3^{5}=243$. When $r=2, s=3$, i.e., $q=\frac{3}{2}, a_{1}=2^{5}$, $a_{6}$ reaches its minimum value 243.