Determine the natural, single-digit numbers $a, b, c$ such that $a, b, \overline{a c}, \overline{c b}$ form a geometric sequence. (The overline indicates that the number is written in decimal notation).
I. solution: According to the problem,

$$
b^{2}=a \cdot \overline{a c}
$$

From this, it is clear that 1) $b$ cannot be a prime number (because if $a=1$, then $b^{2}=10+c$, and there is no one-digit prime number squared between 10 and 20), 2) that $b^{2}$ is a two-digit number, so $b>3$.

Therefore, the possible values of $b$ are: $4,6,8,9$. Their squares are 16, 36, 64, 81, respectively. Breaking these down into the product of a one-digit and a two-digit number, and considering that $a<c$ is necessary, only $1 \cdot 16$ fits the desired form of $a \cdot \overline{a c}$. Thus, $b^{2}=16$, or $b=4, a=1, c=6$.

Indeed, $1,4,16,64$ meet the requirements of the problem.

II. solution. If we denote the common ratio of the geometric sequence by $q$, then

$$
\begin{aligned}
a q & =b \\
a q^{2} & =10 a+c \\
a q^{3} & =10 c+b
\end{aligned}
$$

Subtracting 10 times (2) from (3) and using (1), we get

$$
a q^{3}-10 a q^{2}=a q-100 a
$$

and thus (since $a \neq 0$)

$$
q^{3}-10 q^{2}-q+100=0
$$

From (2), it follows that $q$ can only be positive and from (1), that it can only be rational. From (4), it follows - because the coefficient of $q^{3}$ is 1 - that if $q$ is rational, it can only be an integer, and it can only be chosen from the divisors of 100. We can quickly see that among the values $1,2,4,5$, only

$$
q=4
$$

satisfies (4).

Since (3) states that $a q^{3}=a \cdot 64=10 c+b$, then

$$
a=1, \quad c=6, \quad b=4
$$

Maria Detre (Esztergom, 9th grade mechanical engineering technical school III. class)