1. Three athletes start from the same point on a closed running track that is 400 meters long and run in the same direction. The first runs at a speed of 155 m/min, the second at 200 m/min, and the third at 275 m/min. After what least amount of time will they all be at the same point again? How many overtakes will occur during this time (do not count the start and finish points)?
Answer: $\frac{80}{3}$ minutes $=26$ min. 40 sec.; 13 overtakes.

Solution. The second athlete overtakes the first by 45 m per minute. Therefore, he will catch up with him again after $\frac{400}{45}=\frac{80}{9}$ minutes, meaning the meetings will occur every $\frac{80}{9} n$ minutes ( $n \in N$ ). Similarly, the third athlete overtakes the first by 120 m per minute and will catch up with him again after $\frac{400}{120}=\frac{80}{24}$ minutes, with meetings occurring every $\frac{80}{24} k$ minutes $(k \in N)$. Meetings between the second and third athletes occur at times $\frac{400}{75} m=\frac{80}{15} m(m \in N)$. These three events will coincide if $t=\frac{80}{9} n=\frac{80}{24} k=\frac{80}{15} m$. This will first happen when $n=3, k=8$, and $m=5$ (this can be obtained, for example, by solving the equation $\frac{80}{9} n=\frac{80}{24} k \Leftrightarrow 8 n=3 k \Leftrightarrow n=3 q, k=8 q, q \in Z)$. Thus, $t=\frac{80}{3}$ min.

During this time, the second athlete will overtake the first 2 times (at $n=1$ and $n=2$), the third will overtake the first 7 times, and the second will overtake the third 4 times. The total number of overtakes: $2+7+4=13$.

Answer to variant 172: $\frac{50}{3}$ minutes $=16$ min. 40 sec.; 11 overtakes.

Answer to variant 173: $\frac{80}{3}$ minutes $=26$ min. 40 sec.; 11 overtakes.

Answer to variant 174: 25 minutes; 13 overtakes.