1. In a tennis tournament, 254 schoolchildren are participating. A win earns 1 point, a loss earns 0 points. If the winner initially had fewer points than the opponent, the winner additionally receives one point from the loser. In each round, participants with the same number of points compete, but in one of the pairs, a difference of 1 point between opponents is allowed. The tournament ends as soon as a sole leader is determined. How many schoolchildren will finish the tournament with 5 points?
Answer: 56.

Solution. Let $f(m, k)$ denote the number of students who have $k$ points after $m$ rounds. If $f(m, k)$ is even for any $k \in \{0, \ldots, m\}$, then participants with different numbers of points will not meet in the $(m+1)$-th round (otherwise, there would be at least two such meetings). Also, note that if athletes with $n$ and $n+1$ points meet, they will end up with $n$ and $n+2$ points, respectively, regardless of the outcome. We will prove two statements.

1) In a tournament with $2^n$ participants, after $m$ rounds, $2^{n-m} \cdot C_m^k$ athletes will have $k$ points, where $m \leq n$ and $k \in \{0, \ldots, m\}$. Let $f(m, k)$ be the number of participants who have $k$ points after $m$ rounds. We will use induction on $m$. If $m=0$, then $k=0$, and $f(0,0)=2^n$. For the inductive step, assume that the required equality holds for some $m < n$. Then each group with the same number of points contains an even number of participants, so all participants can be paired. After the $(m+1)$-th round, the groups of athletes with no wins and no losses will be halved. Therefore,

$$
f(m+1,0)=\frac{1}{2} f(m, 0)=\frac{1}{2} \cdot 2^{n-m}=2^{n-(m+1)} \cdot C_{m+1}^{0} \quad \text{and} \quad f(m+1, m+1)=\frac{1}{2} f(m, m)=2^{n-(m+1)} \cdot C_{m+1}^{m+1}
$$

Now, let $k \in \{1, \ldots, m\}$. After the $(m+1)$-th round, $k$ points will be held by those who had $k$ points and lost, as well as those who had $k-1$ points and won. Therefore,

$$
f(m+1, k)=\frac{1}{2} \cdot f(m, k)+\frac{1}{2} \cdot f(m, k-1)=2^{n-m-1}\left(C_m^k + C_m^{k-1}\right)=2^{n-(m+1)} \cdot C_{m+1}^k
$$

(in the last transition, we used the main identity of Pascal's triangle). Thus, the inductive step is complete.
2) In a tournament with $2^n - 2$ participants and fewer than $n$ rounds, the numbers $f(2m, k)$ are even for any $k$ from 0 to $n$, and the numbers $f(2m-1, k)$ are odd only for $k=m-1$ and $k=m$. We will use induction on $n$. For $n=2$, everything is obvious. Assume that the statement is true for some $n$. Then, in a tournament with $2^{n+1}$ participants, after the first round, $2^n - 1$ participants will have 0 points and 1 point. After the second round, $2^{n-1}$ participants will have 0 points and 2 points, and $2^n - 2$ participants will have 1 point. For the first two groups, we use 1), and for the third group, we use the inductive hypothesis.

Now, introduce fictitious participants $A$ and $B$. Let them play against each other in odd rounds, with $A$ winning; this does not affect the other participants. Additionally, let $A$ play against some $A'$ and lose, and $B$ win against some $B'$ in the $(2m)$-th round. Then $A'$ must have $m$ points, and $B'$ must have $(m-1)$ points. By 2), the numbers $f(2m-1, m-1)$ and $f(2m-1, m)$ are odd, and by 1), after adding $A$ and $B$, they become even. Therefore, in the absence of $A$ and $B$, participants $A'$ and $B'$ must play against each other; as a result, they will have $m-1$ and $m+1$ points, respectively, just as after playing against $A$ and $B$. Thus, adding $A$ and $B$ does not affect the distribution of points among the other participants.

Now, apply statement 1) for $n=8$. The tournament will end after 8 rounds, and $C_8^5 = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1} = 56$ participants will have 5 points. Note that $A$ and $B$ will have 4 points each and, therefore, will not be included in this number.