In a single-elimination ping-pong tournament, we know that the winner won more than 68% but less than 69% of their matches. What is the minimum number of participants in the tournament?
In a single-round-robin tournament, every participant plays against every other participant exactly once. Thus, every participant, including the winner, plays $k$ matches, and the number of participants is $k+1$.

Let the number of matches won by the winning participant be $n$. Then, as we know:

$$
0.68 k < n < 0.69 k
$$

From this, by rearranging, we get:

$$
k \geq 15 \text{. }
$$

Since $0.68 \cdot 15 = 10.2$ and $0.69 \cdot 15 = 10.35$, it is impossible for (1) to hold when $k=15$. However, when $k=16$,

$$
0.68 \cdot 16 = 10.88 < 11 < 0.69 \cdot 16 = 11.04
$$

Thus, when $k=16$ and $n=11$, (1)-(6) all hold. However, this is not sufficient to answer the question; it must also be shown that a 17-participant single-round-robin tournament is conceivable where the winner won 11 matches. (Almost all submitters overlooked this.)

# $1984-12-448-1$. eps 

To provide the results of the tournament, consider a participant $V$, who is the winner, and divide the remaining 15 participants into a group of 6 and two groups of 5. $V$ should win against the participants in the group of 6 and one of the groups of 5, and lose to the participants in the other group of 5. This way, $V$ has 11 wins and 5 losses. The participants in the group of 6 should win against those who defeated $V$ and lose to those who were defeated by $V$. Finally, in the two groups of 5, those who defeated $V$ should win against those who were defeated by $V$.

It is not difficult to see that, regardless of the results within the groups, everyone except $V$ has at least 6 losses, making $V$ the true winner.

Thus, at least 17 participants started in the tournament, and 17 could have started.