$A, B, C, D, E$ and $F$ are cities in sequence along a route. Regarding the consecutive straight segments of the route, we know the following: $A B>B C>C D>D E>E F$; the length of each segment is an integer; $A B=2 E F$. The total length of the route from $A$ to $F$ is $53 \mathrm{~km}$. What could be the lengths of the individual segments?
We denote the first and the longest segment as $2d$. Since the segments follow each other in decreasing order and each has an integer measure, the next segment's measure is at most $2d-1$, the following one $2d-2$, and so on.

On the other hand, the smallest segment's measure is $d$ according to the condition. Starting from this, the next segment's length measure is at least $d+1$, the following one at least $d+2$, and so on.

Based on the total length of the $AF=53 \text{ km}$ route, we can write the following inequality for $d$:

$$
2d + (2d-1) + (2d-2) + (2d-3) + d \geq 53 \geq d + (d+1) + (d+2) + (d+3) + 2d
$$

which simplifies to

$$
9d - 6 \geq 53 \geq 5d + 6
$$

from which

$$
6 \frac{5}{9} \leq d \leq 7 \frac{5}{6}
$$

follows, and since $d$ is also an integer, $d = EF = 7$ and $AB = 14$.

Since for $E F B C$

$$
10 + c \leq 13
$$

it follows that

$$
c \leq 3
$$

However, from conditions (1) and (2), $c \geq 2$, otherwise $a + b + c \leq 3$.

If $c = 2$, then $b = 2, a = 1$, so

$$
DE = 9, \quad CD = 11, \quad BC = 12
$$

If $c = 3$, then $b = 2, a = 0$ or $b = 1, a = 1$, and thus

$$
DE = 8, \quad CD = 11, \quad BC = 13
$$

or

$$
DE = 9, \quad CD = 10, \quad BC = 13
$$

According to these, we have found three possibilities for the measures of the road segments.

![](https://cdn.mathpix.com/cropped/2024_05_02_b994a008a51f9810292dg-1.jpg?height=328&width=1219&top_left_y=2080&top_left_x=434)