Problem 3. (4 points) Given a positive number $a$, for which the inequality $1<x a<2$ has exactly three integer solutions. How many integer solutions can the inequality $2<x a<3$ have?
Solution. The first inequality is equivalent to $\frac{1}{a}<x<\frac{2}{a}$. The interval $\left(\frac{1}{a}, \frac{2}{a}\right)$ contains three integers. They divide it into two (internal) intervals of length 1 and two (end) intervals of length not greater than 1. Therefore, for the length $\frac{1}{a}$ of the interval $\left(\frac{1}{a}, \frac{2}{a}\right)$, the inequality $2<\frac{1}{a} \leq 4$ holds.

In general, if an interval of length $t$ contains $k$ integers, then the inequality $k-1<t \leq k+1$ holds, from which $t-1 \leq k<t+1$.

The second interval $\left(\frac{2}{a}, \frac{3}{a}\right)$ has length $t=\frac{1}{a}$, and taking into account the inequalities for $\frac{1}{a}$, we get

$$
1<\frac{1}{a}-1 \leq k<\frac{1}{a}+1 \leq 5
$$

i.e., $k=2,3$ or 4. All three cases are possible: $k=2$ for $a=\frac{3}{8}(x=6$ and 7$); k=3$ for $a=\frac{1}{4}$ $(x=9,10$ and 11$); k=4$ for $a=\frac{5}{17}(x=7,8,9$ and 10$)$.