8. Several buses (more than three) at the beginning of the workday sequentially leave from one point to another at constant and equal speeds. Upon arrival at the final point, each of them, without delay, turns around and heads back in the opposite direction. All buses make the same number of trips back and forth, and the first bus completes its first trip later than the last bus begins its first trip. Each driver counted how many times they met other buses throughout the day, and the total for all drivers was 300. Determine how many buses there were and how many trips they made. In your answer, provide the product of the number of buses and the number of trips.
Answer: 52 or 40.

Solution: If we denote $n-$ as the number of buses and $k-$ as the number of trips, then we can calculate the number of encounters (for example, by a schedule illustrating the movement of buses). It turns out that every two buses meet $2 k-1$ times. In total, we get $n(n-1)(2 k-1)=$ 300, i.e., we need to factorize the number 300 into three factors, one of which is one less than another, and the third is odd. There are two options: $300=4 \times 3 \times 25=5 \times 4 \times 15$. Therefore, $n=4, k=13$ or $n=5$, $k=8$.