92. If the first digit of a three-digit number is increased by $n$, and the second and third digits are decreased by $n$, then the resulting number will be $n$ times the original number. Find the number $n$ and the original number.
Solution. Let $100 x+10 y+z$ be the original number. According to the condition, we have the equality

$$
100(x+n)+10(y-n)+(z-n)=n(100 x+10 y+z)
$$

from which

$$
100 x+10 y+z=\frac{89 n}{n-1}
$$

Since 89 is a prime number, either $n-1$ must be equal to 1, or $n$ must be divisible by $n-1$. In both cases, we arrive at the equality $n=2$, and then the desired number is 178.

Answer: 178.