13.307. Find a three-digit number whose digits form a geometric progression, given that after reducing it by 495, the resulting number is written with the same digits as the desired number but in reverse order; if the digits of the number obtained after subtraction are decreased (from left to right) by 1, 1, and 2, respectively, then an arithmetic progression is formed.
Solution.

Let $100 x+10 y+z$ be the desired number. According to the condition, we have the system

$$
\left\{\begin{array}{l}
x z=y^{2}, \\
100 x+10 y+z-495=100 z+10 y+x, \text { from which } x=9, y=6, z=4 \\
(z-1)+(x-2)=2(y-1)
\end{array}\right.
$$

Answer: 964.