## Task B-3.3.

Determine all three-digit natural numbers $\overline{a b c}$ for which $a+b+c=a \log _{b} c$.
## Solution.

The expression will make sense if $a>0, b>1$ and $c>1$.

In the given expression, the left side is a natural number, so the right side must also be a natural number. For the right side to be a natural number, it must be $c=b^{k}$.

By substituting this equality into the given expression, we get $a+b+b^{k}=a k$.

From $b+b^{k}=a(k-1)$, it follows that $k>1$ because otherwise the left side would be negative, which is impossible. Since $c$ is a natural number between 2 and 9 and $c=b^{k}$, the number $k$ can have the value 2 or 3.

If $k=2$, then

$$
a+b+b^{2}=2 a
$$

$$
b(b+1)=a
$$

Thus, the digit $a$ is the product of two consecutive digits, which is possible only if $b=2$ and $a=6$. Then $c=b^{2}=4$, so it is the number 624.

If $k=3$, then

$$
\begin{gathered}
a+b+b^{3}=3 a \\
b\left(b^{2}+1\right)=2 a
\end{gathered}
$$

If $b=2$, it follows that $a=5$. If the value of $b$ is greater than 2, the value of $a$ will be greater than 9, which is impossible. Therefore, $c=b^{3}=8$, so it is the number 528.

Thus, the sought three-digit numbers are 528 and 624.