7.4. Two swimmers started simultaneously on adjacent lanes in a 50-meter pool for a 1 km distance. Each swam at their own constant speed. The first swimmer, upon completing the swim, got out of the pool. The second continued swimming the distance. In total, the swimmers met 16 times (when the first catches up with the second, it is also a meeting; the start moment is not counted as a meeting). How many times faster can the first swimmer be than the second? List all options and prove that there are no others.
Answer: In $3 / 2.5 / 3$ or 5 times.

Solution: When the first swimmer swam the pool in one direction (which happened 20 times), he met the second swimmer exactly once (possibly at the edge). The meeting at the start is not counted, so there were 19 minus the number of meetings at the edge before the first swimmer finished—such a meeting was counted twice in our calculation. A meeting at the moment the first swimmer finished does not result in a loss.

Thus, during the swim, the swimmers met at the edge three times. Consider the moment of the first such meeting. By this time, the first swimmer had swum some integer number $a$ of pool lengths, and the second swimmer had swum some integer number $b$. Then the speeds of the swimmers are in the ratio $a / b$. Simplify this fraction and assume that the speeds are in the ratio $p / q$.

If $p$ and $q$ are both odd, then when the first swimmer swims $p$ pool lengths, the second swimmer is at the same edge, meaning a meeting occurs, and this repeats at such intervals. Therefore, $3 p < 20$ (the first swimmer could not have finished after three meetings, as $3 p$ is odd) and $20 \leq 4 p$, since there were at most 4 meetings, including the moment the first swimmer finished. Thus, $p=5$, and $q=1$ or $q=3$. It is not difficult to verify that both fractions $5 / 1$ and $5 / 3$ are suitable.

If $p$ and $q$ have different parities, then at the moment the first swimmer swims $p$ pool lengths, they are at opposite edges, and a meeting at the edge will occur when the first swimmer swims $2 p, 4 p, 6 p \ldots$ pool lengths. Therefore, to have only three meetings, $6 p \leq 20 < 8 p$, from which $p=3, q=2$.

## Criteria:

Only the answer - 0 points.

Answer with verification, no other progress - 1 point. This point does not accumulate with others.

Proved that the swimmers met at the edge three times - 1 point.

Proved that the speeds are in the ratio of a fraction - another 1 point.

Correctly considered one of the cases in the solution - another 3 points.

Correctly considered the second - another 2 points.

No points should be deducted for the lack of verification that the ratios are suitable.