2. In order to find out the speed at which the bus he was riding in was moving, a passenger looked out the window and saw a two-digit number on the kilometer marker by the road (which indicates the distance from the city of departure). Exactly one hour later, he looked out the window again and saw a three-digit number on the kilometer marker, written with the same two digits as an hour before, but in reverse order with a zero in between. The passenger calculated the speed of the bus and fell asleep. When he woke up exactly two hours later and looked out the window, he saw a three-digit number on the kilometer marker where the first and last digits were the same as those of the number he saw two hours ago, but the middle digit was different from zero. If it is known that the bus was moving at a constant speed, determine the speed of the bus and the numbers the passenger saw on the kilometer markers by the road.
Solution. The first time the traveler saw the number $10 x+y, 0<x \leq 9$, $0 \leq y \leq 9$. After one hour, he saw the number $100 y+x, y \neq 0$. During this time, the bus traveled $(100 y+x)-(10 x+y)=9(11 y-x) \text{ km}$. Since the bus was moving at a constant speed, in the next two hours, the bus traveled $2 \cdot 9(11 y-x)=18(11 y-x)$, and the traveler saw the number $100 y+10 z+x, 0<z \leq 9$. We get

$$
18(11 y-x)=(100 y+10 z+x)-(100 y+x) \text{, i.e., } 9(11 y-x)=5 z
$$

The last equation is possible only if $z=9$ and $11 y-x=5$. From $0<x, y \leq 9$ we get $x=6$ and $y=1$. Thus, the speed $v$ of the bus is obtained from $9(11 y-x)=1 \cdot v$, from which $v=45 \text{ km/h}$, and the numbers the traveler saw on the kilometer markers are 61, 106, and 196.