We are traveling from Budapest to Košice by car at a constant speed. We see a kilometer marker with a two-digit number. Half an hour later, we reach a sign where the digits are in reverse order. Another 30 minutes later, we reach a sign that has the two previous digits plus a 0. How fast are we traveling?
Let the first two-digit number be $\overline{a b}$, half an hour later the number on the kilometer sign is $\overline{b a}$, and then another half hour later, one of the three-digit numbers $\overline{a 0 b}, \overline{a b 0}$, $\overline{b 0 a}, \overline{b a 0}$ appears, since the first digit cannot be 0.

Let this three-digit number be denoted by $A$. Since we traveled at a constant speed, that is, we covered equal distances in equal times, we can write the following equation:

$$
(10 b+a)-(10 a+b)=A-(10 b+a)
$$

Rearranging the equation:

$$
A=19 b-8 a
$$

Since the value of $b$ can be at most 9, and $a$ is at least 1, we have

$$
A \leq 19 \cdot 9-8=163
$$

so the first digit of $A$ is 1.

Since the numbers on the kilometer sign are increasing, $\overline{a b}$ is definitely smaller than $\overline{b a}$, so the value of $a$ is 1 and the three-digit number is either $\overline{1 b 0}$ or $\overline{10 b}$.

If $A=100+10 b$, then from equation (1) we get

$$
100+10 b=19 b-8
$$

from which $9 b=108$ and $b=12$, which is impossible.

If $A=100+b$, then from equation (1) we get $100+b=19 b-8$, from which $18 b=108$ and $b=6$.

Therefore, the three-digit number is 106.

The speed of the car is therefore $\frac{106 \mathrm{~km}-61 \mathrm{~km}}{\frac{1}{2} \text { hour }}=90 \frac{\mathrm{km}}{\text { hour }}$.

Balázs Kecskeméti (Szeged, Ságvári E. Gymnasium, 12th grade)