15. In a geometric sequence with a common ratio greater than 1, what is the maximum number of terms that are integers between 100 and 1000?
15. $128,192,288,432,648,972$ are integers between 100 and 1000, forming a geometric sequence with a common ratio of $\frac{3}{2}$. There are 6 terms in total.

Let the geometric sequence $\left\{a r^{n-1}\right\}$ satisfy $100 \leqslant a$, and it is clear that $r$ is a rational number. Let $r=\frac{p}{q}, p>q \geqslant 1, p, q \in \mathbf{N},(p, q)=1$.
Since $a r^{n-1}=a\left(\frac{p}{q}\right)^{n-1}$ is an integer and $(p, q)=1$, it follows that $q^{n-1}$ divides $a$.
First, we take $p=q+1$, and then determine the number of terms $n$ in the integer sequence $a<a\left(\frac{q+1}{q}\right)<\cdots<a\left(\frac{q+1}{q}\right)^{n-1}$ within the range $[100,1000]$:
If $q \geqslant 3$, then $1000 \geqslant a\left(\frac{q+1}{q}\right)^{n-1} \geqslant(q+1)^{n-1} \geqslant 4^{n-1}$, in this case, $n \leqslant 5$;
If $q=1$, then $1000 \geqslant a\left(\frac{q+1}{q}\right)^{n-1}=a \cdot 2^{n-1} \geqslant 100 \cdot 2^{n-1}$, in this case, $n \leqslant 4$;
If $q=2$, then $1000 \geqslant a\left(\frac{q+1}{q}\right)^{n-1}=a \cdot\left(\frac{3}{2}\right)^{n-1} \geqslant 100 \cdot\left(\frac{3}{2}\right)^{n-1}$, in this case, $n \leqslant 6$.
Secondly, if $p \geqslant q+2$, similarly, we can get that the number of terms $n$ in the integer sequence $a<a\left(\frac{q+1}{q}\right)<\cdots<a\left(\frac{q+1}{q}\right)^{n-1}$ within the range $[100,1000]$ is $n \leqslant 5$.
In summary, in a geometric sequence with a common ratio greater than 1, the maximum number of terms that are integers between 100 and 1000 is 6.