2. Misha invited eighteen friends from his sports club and two of his brothers to celebrate his birthday, a total of twenty guests. All the guests and Misha himself, seated at two tables, ate all the hot dogs served equally on both tables, and everyone ate only from their own table. Each friend from the sports club ate more hot dogs than each of Misha's brothers, but less than Misha by the same number of pieces. How many friends from the sports club and how many brothers sat at the same table with Misha?
# Answer: 9 friends from the sports club and no brothers.

Solution. Let $x$ be the number of brothers and $y$ be the number of friends from the sports club sitting at the same table with Misha. Then, at the other table, there were $2-x$ brothers and $18-y$ friends from the sports club.

Let $c$ be the number of hot dogs each friend from the sports club ate, and $s$ be the constant difference in hot dogs mentioned in the problem.

Then, according to the problem's condition, we can write

$$
\begin{gathered}
x(c-s)+y c+(c+s)=(2-x)(c-s)+(18-y) c \\
(2 x+2 y-19) c=(2 x-3) s .
\end{gathered}
$$

The number $x$ can take the values 0, 1, or 2. Let's consider each of them in turn.

1) $x=0 \Rightarrow(19-2 y) c=3 s$. Since $0<s<c$, then $0<(19-2 y) c<3 c \Rightarrow$

$$
0<(19-2 y)<3 \Rightarrow y=19-\text { a single integer }
$$

2) $x=1 \Rightarrow(17-2 y) c=s$. Since $0<s<c$, then $0<(17-2 y) c<$ c $\Rightarrow$ $0<(17-2 y)<1 \Rightarrow y-$ no integers.
3) $x=2 \Rightarrow(2 y-15) c=s$. Since $0<s<c$, then $0<(2 y-15) c<$ c $\Rightarrow$

$$
0<(2 y-15)<1 \Rightarrow y-\text { no integers. }
$$

Thus, the situation described in the problem can only occur when $x=0, y=9$.