8.2. In the decimal representation of a natural number $N$, a 0 was inserted between the second and third digits from the right, resulting in $9N$. What can $N$ be? (list all possible values).
Answer: 225; 450; 675.

Solution. Let $x$ be the third digit from the right of the number $N$. The two-digit number formed by the last two digits of the number $N$ will be denoted by $B$, and the number obtained from $N$ by erasing its last three digits will be denoted by $A$. Thus, $N=1000 A+100 x+B$. The new number, obtained after inserting the digit 0, will be equal to $10000 A+1000 x+B$. From the conditions of the problem, we get the equation $10000 A+1000 x+B=9000 A+900 x+9 B$, i.e., $1000 A+100 x=8 B$. If $A \neq 0$, then the number on the left side of the last equation is at least 1000, while the number on the right is less than 800. Therefore, $A=0$ and the equation simplifies: $25 x=2 B$. Thus, $B$ is a two-digit number divisible by 25 (if $B=0$, the number $N=0$ is not a natural number). Therefore, there are three possible cases: either $B=25$, or $B=50$, or $B=75$. The corresponding digit $x$ is 2 or 4 or 6.