## Task 11/73

A train passes a kilometer post with a two-digit kilometer number. After the time $\Delta t_{1}$, it passes another kilometer post with the same digits but in reversed order.

Finally, after the additional time $\Delta t_{2}=\Delta t_{2}$, it encounters a third kilometer post, whose number is the same as the first number with a zero inserted in between.

The average speeds $v_{1}$ and $v_{2}$ during the times $\Delta t_{1}$ and $\Delta t_{2}$, respectively, are the same: $v_{1}=v_{2}$.

What are the numbers on the kilometer posts?
Given $v_{1}=v_{2}$ and $\Delta t_{1}=\Delta t_{2}$, it follows that $\Delta s_{1}=v_{1} \Delta t_{1}=v_{2} \Delta t_{2}=\Delta s_{2}$.

This means the two segments between the three milestones are equal. Therefore, the numbers on the milestones form an arithmetic sequence of the first order.

If we denote the first number as $10 a+b$ $(1 \leq a \leq 9, 0 \leq b \leq 9$, where $a$ and $b$ are integers), then according to the problem, the second number is $10 b+a$ (which immediately implies $b \neq 0$) and the third number is $100 a+b$. It follows that

$$
100 a+b-10 b-a=10 b+a-10 b-b \rightarrow b=6 a
$$

Given the constraints on $a$ and $b$, the only solution to this Diophantine equation is $a=1, b=6$. Therefore, the numbers on the milestones are 16, 61, 106 (the difference is 45).