1. In an arm wrestling tournament, 510 athletes are participating. 1 point is awarded for a win, and 0 points for a loss. If the winner initially had fewer points than the opponent, the winner additionally receives one point from the loser. In each round, participants with a difference of no more than 1 point in their scores compete. The tournament ends as soon as a sole leader is determined. What is the minimum number of rounds that need to be played?
# Answer: 9.

Solution. Let there be $N=2 m+k$ athletes in the leading group before the next round, with $2 m$ of them meeting each other, and $k$ meeting participants who have one point less. Note that if athletes with $n$ and $n+1$ points meet, they will end up with $n$ and $n+2$ points regardless of the outcome. Therefore, after the round, there will be $m+k$ participants in the leading group. The minimum value of this sum is $\frac{N}{2}$ for even $N$ and $\frac{N+1}{2}$ for odd $N$. According to the problem, after the first round, there will be no fewer than 255 leaders, after the second round - no fewer than 128, and after each subsequent round, the number of leaders will decrease by no more than half. Therefore, it will take no fewer than $2+\log _{2} 128=9$ rounds.

We will show that the tournament can be completed in 9 rounds. We will prove two statements.

1) In a tournament with $2^{n}$ participants, the winner can be determined in $n$ rounds. We will use induction on $n$. For $n=1$, this is obvious. Suppose the statement is true for some $n$. Then, in a tournament with $2^{n+1}$ participants, after the first round, $2^{n}$ athletes will have 0 points and $2^{n}$ will have 1 point. We will conduct two independent tournaments in these groups. By the induction hypothesis, both can be completed in $n$ rounds. The winner of the second tournament will become the overall winner, and he will be determined in $n+1$ rounds.
2) For $n \geqslant 3$, in a tournament with $2^{n}-2$ participants, the winner can be determined in $n$ rounds. After the first round, $2^{n-1}-1$ athletes will have 0 points and $2^{n-1}-1$ will have 1 point. If in the second round, only one pair of opponents has a different number of points, then after the round, $2^{n-2}$ participants will have 0 points and 2 points, and $2^{n-1}-2$ participants will have 1 point. We will now use induction on $n$. For $n=3$, we will pair athletes with the same number of points, after which the tournament will end. Suppose $n>3$ and the statement is true for $n-1$. By statement 1) and the induction hypothesis, independent tournaments can be organized in the groups with 0, 1, and 2 points, and the first and third groups will end in $n-2$ rounds, while the second group will end in $n-1$ rounds (but the last round will not be needed). The winner of the third group will win the entire tournament, as he will have $n+1$ points, while participants from other groups will have no more than $n$ points. In total, the tournament will last $n$ rounds.

It remains to apply statement 2) for $n=9$.