2. What is the maximum possible number of consecutive terms of an increasing geometric progression that can be three-digit natural numbers? Provide an example of such a sequence. (16 points)
Solution. Let the required members of the progression be $a_{0}, a_{1}, \ldots, a_{n}, a_{k}=a_{0} q^{k}$, the common ratio - an irreducible fraction $q=r / s, r>s$. Then $a_{0}=b s^{n}, a_{n}=b r^{n}, b \in \mathbb{N}$, since $r^{n}$ and $s^{n}$ are coprime. We obtain the restriction

$$
r^{n}<1000 / b, \quad s^{n} \geq 100 / b
$$

1) Let $r=2$, then $q=2 / 1, (2) \Longrightarrow n \leq 9$. For $n=9$ we have

$$
r=2, \quad s=1, \quad\left\{\begin{array}{c}
a_{0}=b \geq 100 \\
a_{n}=512 b<1000
\end{array} \Longrightarrow b=1\right.
$$

However, $a_{9}=512$ is not a valid solution since $a_{9} \geq 1000$ is required. Thus, $r=2$ is not useful.

2) Let $r=3$, then $q=3 / 2, (2) \Longrightarrow n \leq 5$. For $n=5$ we have

$$
r=3, \quad s=2, \quad\left\{\begin{array}{c}
a_{0}=32 b \geq 100 \\
a_{n}=243 b<1000
\end{array} \Longrightarrow b=4\right.
$$

3) Let $r \geq 4$, then ( 1$) \Longrightarrow n \leq 4$. The record $n=5$ cannot be broken.

Answer: 6. The only example: $128,192,288,432,648,972$.