13.375 Find a three-digit number, knowing that the number of its tens is the geometric mean of the number of hundreds and units. If in its notation the digits of hundreds and units are swapped and the new number is subtracted from the desired one, then the difference will be equal to 297.
Solution. Let $\overline{x y z}$ be the desired three-digit number. Then $\overline{x y z}-\overline{z y x}=297$ $\Rightarrow(100 x+10 y+z)-(100 z+10 y+x)=297 \Rightarrow x=z+3$.

By the condition, $y^{2}=x z=z(z+3) \Rightarrow y$ is even $\Rightarrow z(z+3) \vdots 4 \Rightarrow$

$\Rightarrow\left[\begin{array}{l}z: 4, \\ (z+3): 4\end{array} \Rightarrow\left[\begin{array}{l}z=0, \\ z=4, \\ z+3=4, \\ z+3=8 .\end{array}\right.\right.$ Considering now that $y^{2}=z(z+3)$, we find that $z=0$

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-332.jpg?height=106&width=408&top_left_y=625&top_left_x=87)

Answer: 300 or 421.