If we write the number $F$ in the base-$a$ number system, we get the repeating decimal $0.3737 \ldots=0. \dot{3} \dot{7}$, and for the number $G$, we get $0.7373 \ldots=0. \dot{7} \dot{3}$.

These same numbers, when written in the base-$b$ number system, are $F=0.2525 \ldots=0.2 \dot{2}$ and $G=0.5252 \ldots=0.5 \dot{2}$. Determine the values of $a$ and $b$.
Solution. It is known that in a number system with a given base, numbers can be multiplied by the base of the number system by shifting the comma, which is the decimal point in the decimal system, one place to the right. Accordingly, the numbers $a \cdot F$ and $a \cdot G$ in the base-$a$ number system are

$$
a \cdot F=3.7373 \ldots=3. \dot{7} \dot{3} \quad \text { and } \quad a \cdot G=7.3737 \ldots=7. \dot{3} \dot{7}
$$

Since the fractional parts of $a \cdot F$ and $G$ are equal, their difference is an integer:

$$
a \cdot F-G=3
$$

Similarly, we get

$$
a \cdot G-F=7
$$

When the numbers written in the base-$b$ number system are multiplied by $b$, we similarly obtain

$$
\begin{aligned}
& b \cdot F-G=2. \dot{5} \dot{2}-0. \dot{5} \dot{2}=2, \quad \text { and } \\
& b \cdot G-F=5. \dot{2} \dot{5}-0. \dot{2} \dot{5}=5
\end{aligned}
$$

From the resulting system of four equations, we need to determine the values of $a$ and $b$, knowing that both are integers greater than 1.

Subtracting equation (1) from equation (3) gives $(a-b) F=1$, and subtracting equation (2) from equation (4) gives $(a-b) G=2$. Therefore, $G=2 F$. From equations (1) and (2), we have

$$
\frac{1}{F}=\frac{a-2}{3}=\frac{2 a-1}{7}
$$

The value of $a$ from this equation is 11. Similarly, from equations (3) and (4), we get

$$
\frac{1}{F}=\frac{b-2}{2}=\frac{2 b-1}{5}
$$

from which $b=8$.

Thus, the solution to the problem is $a=11$ and $b=8$. In this case, $F=\frac{1}{3}$ and $G=\frac{2}{3}$, and these fractions are indeed of the given form in the base-11 and base-8 number systems, respectively.