7. In the school Spartakiad, teams from classes $8^{\mathrm{A}}$, $8^{\text {Б }}$, and $8^{\mathrm{B}}$ participated. In each of the competitions, one of these teams took 1st place, another took 2nd place, and another took 3rd place. After the Spartakiad, points were tallied: $x$ points were awarded for 1st place, $y$ for 2nd place, and $z$ for 3rd place $\left(x>y>z>0-\right.$ integers). In the end, team $8^{\mathrm{A}}$ received 22 points, and teams $8^{\text {Б }}$ and $8^{\text {В }}$ each received 9 points. How many competitions were there in total, and which team took 2nd place in the grenade throwing competition, given that the team $8^{\text {B }}$ took 1st place in the vault over the "goat"?
Answer: 5 competitions, $8^{\text {B }}$.

Solution: Let $n \geqslant 2$ be the number of competitions in the Spartakiad, then the total number of points scored by all teams is $n(x+y+z)=22+9+9=40$. But $z \geqslant 1, y \geqslant 2, x \geqslant 3$, so $x+y+z \geqslant 6$. Consider the possible cases: $x+y+z=8, n=5$; $x+y+z=10, n=4$ and $x+y+z=20, n=2$.

(a) Case $x+y+z=10, n=4$. Clearly, $x \leqslant 6$ (otherwise $8^{\mathrm{B}}$ would score more than 9 points). Consider the possible cases:

- $x=6$. Then $y=3, z=1$, but then $8^{\mathrm{A}}$ would not score 22 points.
- $x=5$. Then $8^{\mathrm{A}}$ would score less than 20 points.
- $x \leqslant 4$. Then $x+y+z \leqslant 9$.

Thus, the specified case is impossible.

(b) Case $x+y+z=20, n=2$. If we assume that $x<11$, then team $8^{\mathrm{A}}$ would not be able to score 22 points. If $x \geqslant 11$, then team $8^{\text {B }}$ would end up with more than 11 points, which is incorrect. Thus, the specified case is impossible.

(c) Case $x+y+z=8, n=5$. Clearly, in this case $z=1$, $x+y=7$. Consider the cases:

- Suppose $x=4, y=3$. Then $y+z=4$, so teams $8^{\text {B }}$ and $8^{\text {B }}$ scored at least 4 points in each competition. Then over 5 competitions, they should have scored at least 20 points (while in fact they scored $9+9=18$ ).
- Suppose $x=5, y=2$. Then team $8^{\text {B }}$ once took 1st place and 4 times last. Team $8^{\mathrm{A}}$ did not take first place in the vault. The only way to score 22 points: 2nd place and victory in the remaining 4 competitions. This means that team $8^{\text {B }}$ in these four competitions consistently took 2nd place (and in the vault $-3-\mathrm{rd})$.