Problem 8.5. A car left the city for the village at the same time a cyclist left the village for the city. When the car and the cyclist met, the car immediately turned around and headed back to the city. As a result, the cyclist arrived in the city 35 minutes after the car. How many minutes did the cyclist spend on the entire journey, given that his speed is 4.5 times slower than the speed of the car?
Answer: 55.

![](https://cdn.mathpix.com/cropped/2024_05_06_b135fe79f2e6ed0cbffag-3.jpg?height=113&width=493&top_left_y=1217&top_left_x=478)

Fig. 7: to the solution of problem 8.5

Solution. In Fig. 7, let's mark village $A$, city $B$, point $P$ where the car and the cyclist meet, and point $Q$, where the cyclist was when the car returned to the city. Since the speeds of the car and the cyclist differ by 4.5 times, then $A P: P B=1: 4.5=2: 9$. Since the car spent the same amount of time on the movements $B \rightarrow P$ and $P \rightarrow B$, the cyclist also spent the same amount of time on the corresponding movements $A \rightarrow P$ and $P \rightarrow Q$. Therefore, $A P: P Q: Q B=2: 2: 7$.

Since the cyclist spent exactly 35 minutes on the movement $Q \rightarrow B$, the total time spent on the movement $A \rightarrow B$ is proportional: $35 \cdot \frac{11}{7}=55$ minutes.

Pasha listed all natural divisors of a natural number $k$ in ascending order and numbered them: the first, the second,

Pasha noticed that if the sixth divisor is multiplied by the thirteenth divisor, the result is the original number $k$.

How many natural divisors does the number $k$ have?

Answer: 18.

Solution. Let the natural divisors of the number $k$ be ordered as follows:

$$
1=d_{1}<d_{2}<\ldots<d_{6}<\ldots<d_{13}<\ldots<d_{m-1}<d_{m}=k
$$

The numbers

$$
1=\frac{k}{d_{m}}<\frac{k}{d_{m-1}}<\ldots<\frac{k}{d_{6}}<\ldots<\frac{k}{d_{13}}<\ldots<\frac{k}{d_{2}}<\frac{k}{d_{1}}=k
$$

are also divisors of the number $k$, they are distinct, and there are as many of them. Thus, these are the same numbers, only in reverse order. We get that

$$
d_{1}=\frac{k}{d_{m}}, d_{2}=\frac{k}{d_{m-1}}, \ldots, \quad d_{m}=\frac{k}{d_{1}}
$$

Thus, the divisors are divided into pairs of "opposite" divisors, which give the original number $k$ when multiplied:

$$
k=d_{1} \cdot d_{m}=d_{2} \cdot d_{m-1}=\ldots
$$

In each such pair, the sum of the indices of the divisors is $m+1$. Since by the condition $d_{6} \cdot d_{13}=k$, we get that $m=6+13-1=18$.