In a one-off table tennis tournament, a) 10, b) 11 players started. How many matches did the champion win at least, given that there was no tie for first place?
Let's consider the general case: let the number of participants be $n$, and the champion wins $k$ matches. There was no tie for the lead, so each of the others won at most $(k-1)$ times. The number of matches, counted from the winners' perspective, is therefore at most $k+(n-1)(k-1)$. In a round-robin tournament with $n$ participants, where everyone plays everyone else, the number of matches is $\frac{n(n-1)}{2}$. This means that

$$
\begin{gathered}
\frac{n(n-1)}{2} \leqq k+(n-1)(k-1), \text { hence } \\
\frac{n+1}{2}-\frac{1}{n} \leq k
\end{gathered}
$$

The smallest integer $k$ for which (2) holds is $\left[\frac{n}{2}\right]+1$, so the champion must win at least $\left[\frac{n}{2}\right]+1$ matches. This can indeed happen, for example, as follows: Let the participants stand in a circle, and each player defeat the $\left[\frac{n}{2}\right]$ players standing to their left, if they haven't already defeated them. This way, everyone plays everyone else, each player wins at most $\left[\frac{n}{2}\right]$ matches, and there is one, for example, the one to whom we first assigned the defeated players, who wins exactly that many. Change one of their lost matches to a win. This way, they become the champion with $\left[\frac{n}{2}\right]+1$ wins.

According to the above reasoning, for $n=10$ and $n=11$, the tournament winner must have won at least 6 matches, and there is indeed a possible outcome of the tournament where 6 wins are sufficient for the first place.

Remarks. 1. Many did not prove that it is possible for the tournament to have an outcome where someone wins with 6 victories, i.e., that the lower bound for the number of the champion's wins is sharp.

2. It can be seen that if ties for the lead are allowed, then 5 wins may be sufficient to win the tournament.