11. Given a geometric sequence $\left\{a_{n}\right\}$ with a common ratio greater than 1, each term is an integer, and each term satisfies $100 \leqslant a_{n} \leqslant$ 1000, how many terms can the sequence $\left\{a_{n}\right\}$ have at most? Write the sequence $\left\{a_{n}\right\}$ with the maximum number of terms (require writing every term of the sequence).
11. Solution: Given that each term of the geometric sequence $\left\{a_{n}\right\}$ is an integer, and the common ratio is greater than 1, it is easy to know that the common ratio $r$ is a rational number. Let $r=\frac{p}{q}$, where $p>q \geqslant 1, p, q \in \mathbf{N}^{*}$, and $p, q$ are coprime.
Then we have $a_{n}=a_{1}\left(\frac{p}{q}\right)^{n-1} \geqslant a_{1}\left(\frac{q+1}{q}\right)^{n-1}$, and $a_{1} \geqslant q^{n-1}$.
(1) If $q=1$, since $a_{1} \geqslant 100$, then $100 \cdot 2^{n-1} \leqslant a_{1} \cdot 2^{n-1} \leqslant a_{n} \leqslant 1000$, then $2^{n-1} \leqslant 10$, so $n \leqslant 4$.
(2) If $q=2$, since $a_{1} \geqslant 100$, then $100 \cdot\left(\frac{3}{2}\right)^{n-1} \leqslant a_{1} \cdot\left(\frac{3}{2}\right)^{n-1} \leqslant a_{n} \leqslant 1000$, then $\left(\frac{3}{2}\right)^{n-1} \leqslant 10$, so $n \leqslant 6$.
(3) If $q \geqslant 3$, since $a_{1} \geqslant q^{n-1}$, then $1000 \geqslant a_{n} \geqslant a_{1}\left(\frac{q+1}{q}\right)^{n-1} \geqslant(q+1)^{n-1} \geqslant 4^{n-1}$, i.e., $4^{n-1} \leqslant 1000$, so $n \leqslant 5$.

In summary, the number of terms in the sequence that meet the conditions does not exceed 6. Below, we find a geometric sequence $\left\{a_{n}\right\}$ with 6 terms. From the above, we know that the common ratio $r=\frac{3}{2}$ at this time. According to each term being an integer, and $a_{6}=a_{1}\left(\frac{3}{2}\right)^{5} \leqslant 1000$, and $a_{1} \geqslant 100$, so $a_{1}=$ $2^{7}=128$, thus the sequence is $128,192,288,432$, 648,972.