1. In an arm wrestling tournament, there are $2^{n}+6$ athletes participating, where $n$ is a natural number greater than 7. One point is awarded for a win, and zero points for a loss. Before each round, pairs are formed by drawing lots among participants with the same number of points (those who do not find a pair skip the round). How many athletes will have 4 points after the seventh round?
Answer: $35 \cdot 2^{n-7}+2$.

Solution. First, we prove two statements.

1) In a tournament with $2^{n}$ participants, after the $m$-th round, the number of athletes with $k$ points will be $2^{n-m} \cdot C_{m}^{k}$, where $m \leqslant n$ and $k \in\{0, \ldots, m\}$. Let $f(m, k)$ be the number of participants who have scored $k$ points after $m$ rounds. We use induction on $m$. If $m=0$, then $k=0$, and $f(0,0)=2^{n}$. We proceed with the inductive step. Suppose for some $m<n$ the required equality holds. 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 scored 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 completed.

2) The tournament can be reduced to two independent tournaments in groups of $2^{n}$ and 6 people. Indeed, in each round, we will only pair athletes from the same group. By 1), all participants in the first group can be paired. Therefore, if someone is left without a pair in the overall tournament, we can consider this participant to be from the second group. At the end of the tournament, the results of the two groups are simply combined.

We calculate the total number of participants who scored 5 points after the seventh round. By 1), in the first group, there will be $2^{n-7} \cdot C_{7}^{4}=35 \cdot 2^{n-7}$. The dynamics of the distribution of points among participants from the second group up to the seventh round are as follows:

$$
6 \rightarrow(3,3) \rightarrow(2,3,1) \rightarrow(1,3,2) \rightarrow(1,2,2,1) \rightarrow(1,1,2,2) \rightarrow(1,1,1,2,1) \rightarrow(1,1,1,1,2)
$$

(the left number represents the number of underdogs, the right number represents the number of leaders). Thus, after the seventh round, two athletes will have 4 points. Therefore, the answer is the sum of $35 \cdot 2^{n-7}$ and 2.