A three-digit number written with different digits has its middle digit equal to the arithmetic mean of the outer digits. Write the last digit before the first two and form the geometric mean of the thus obtained number and the original number. This way, we get the number that results from the original number by moving the first digit to the end of the other two. Determine the digits. - Are there digit triples with such properties in other number systems as well?
I. solution. I. Denoting the digits of the number by $A, B, C$, according to the conditions,

\[
\begin{gathered}
2 B = A + C \\
\left(10^{2} A + 10 B + C\right)\left(10^{2} C + 10 A + B\right) = \left(10^{2} B + 10 C + A\right)^{2}
\end{gathered}
\]

After expanding and rearranging (2), we can simplify by $10^{3}-1$:

\[
A^{2} - 10 B^{2} + 10 A C - B C = 0
\]

From here, eliminating $B$ using (1):

\[
A^{2} - 3 A C + 2 C^{2} = 0
\]

Clearly, $C \neq 0$, otherwise $A=0$, then $B=0$ would follow. Moreover, the sum of the coefficients $1-3+2=0$, so with $A=C$ the equation holds, meaning the left side is divisible by the difference $A-C$. After factoring:

\[
(A-C)(A-2 C) = 0
\]

This can only hold if $A-2 C=0, A=2 C$, because $A$ and $C$ are assumed to be different, $A-C \neq 0$. Thus, from (1) $B=3 C / 2$.

Accordingly, $C$ is an even digit, and $A=2 C \leq 9$, because $A$ is also a digit, so $C \leq 4.5$. Therefore, $C$ can be 2 or 4, and $B$ must be an integer. For $C=2$, $B=3$ and $A=4$. For $C=4$, $B=6$ and $A=8$.

II. solution. (for any base system). We can reach the result more quickly if we write the condition (1) as

\[
A - B = B - C = d, \quad \text{that is} \quad A = B + d, C = B - d
\]

where $d$ is an integer. Then, in the base $b$ number system $(b \geq 3)$, equation (2) becomes:

\[
\begin{gathered}
{\left[(B+d) b^{2} + B b + B - d\right]\left[(B-d) b^{2} + (B+d) b + B\right]} \\
= \left[B b^{2} + (B-d) b + (B+d)\right]^{2} \\
d\left(b^{3}-1\right)[-d b + 3 B + d] = 0
\end{gathered}
\]

From here, for $d \neq 0$ and $b \neq 1$,

\[
B = \frac{(b-1) d}{3}, \quad A = \frac{(b+2) d}{3}, \quad C = \frac{(b-4) d}{3}
\]

Since $B \geq 0$, $d > 0$, and $A > B > C$. Moreover, since $A \leq b-1$, $d \leq 2$, and $d$ cannot be divisible by 3, we only get integer solutions for $A, B, C$ if $b-1 = 3k$. For $b=4$ and $d=2$, $A > b-1$ is not a solution. For all other valid $b$, there are two solutions.

For $b=4$ and $d=2$:

\[
A = \frac{(4+2) \cdot 2}{3} = 4, \quad B = \frac{(4-1) \cdot 2}{3} = 2, \quad C = \frac{(4-4) \cdot 2}{3} = 0
\]

Thus, the number in base 4 is $420_4 = 36_{10}$, and the properties hold:

\[
36 \cdot 9 = 18^2
\]

For $b=7$ and $d=1$:

\[
A = \frac{(7+2) \cdot 1}{3} = 3, \quad B = \frac{(7-1) \cdot 1}{3} = 2, \quad C = \frac{(7-4) \cdot 1}{3} = 1
\]

Thus, the number in base 7 is $321_7 = 162_{10}$, and the properties hold:

\[
162 \cdot 72 = 108^2
\]

For $b=7$ and $d=2$:

\[
A = \frac{(7+2) \cdot 2}{3} = 6, \quad B = \frac{(7-1) \cdot 2}{3} = 4, \quad C = \frac{(7-4) \cdot 2}{3} = 2
\]

Thus, the number in base 7 is $642_7 = 330_{10}$, and the properties hold:

\[
330 \cdot 180 = 252^2
\]