6.30 There is a sports competition consisting of $M$ events, with athletes $A, B, C$ participating. In each event, the first, second, and third places receive $p_{1}, p_{2}, p_{3}$ points, respectively, where $p_{1}, p_{2}, p_{3}$ are positive integers, and $p_{1}>p_{2}>p_{3}$. In the end, $A$ scores 22 points, $B$ and $C$ both score 9 points, and $B$ comes in first in the 100-meter dash.
Find the value of $M$, and determine who came in second in the high jump.
[Solution] Considering the total score of each person, we have
$$
M\left(p_{1}+p_{2}+p_{3}\right)=22+9+9=40.
$$

Given that $p_{1}, p_{2}, p_{3}$ are positive integers, and $p_{1}>p_{2}>p_{3}$, we get
$$
p_{1}+p_{2}+p_{3} \geqslant 1+2+3=6,
$$

thus
$$
\begin{array}{c}
6 M \leqslant 40, \\
M \leqslant 6.
\end{array}
$$

Also, from the problem statement, there are at least the 100-meter dash and high jump, so
$$
M \geqslant 2.
$$

Furthermore, $M$ should be a divisor of 40.
Therefore, $M$ can only take the values $2, 4, 5$.
(1) When $M=2$, we look at $B$'s total score of 9, then
$$
9 \geqslant p_{1}+p_{3},
$$

thus
$$
p_{1} \leqslant 8.
$$

Therefore, if $A$ wins two first places, then $A$'s total score will not exceed 16, which is impossible to reach 22 points, so $M \neq 2$.
(2) When $M=4$, we still look at $B$'s total score of 9, then
$$
9 \geqslant p_{1}+3 p_{3},
$$

Given that $p_{3} \geqslant 1$, so $p_{1} \leqslant 6$.
If $p_{1} \leqslant 5$, then winning first place in all four events would score at most 20 points, which contradicts $A$ scoring 22 points.
So $p_{1}=6$, at this point we have
$$
\begin{array}{c}
4\left(6+p_{2}+p_{3}\right)=40, \\
p_{2}+p_{3}=4.
\end{array}
$$

Since $p_{2} \neq p_{3}$, so $p_{2}=3, p_{3}=1$.
Since $B$ has already won the 100-meter dash, $A$ can win at most 3 first places, at this point $A$'s maximum score is
$$
3 p_{1}+p_{2}=18+3=21.
$$

This contradicts $p_{2}>p_{3}$, so
$$
p_{1}+p_{2}+p_{3} \geqslant 4+3+2=9.
$$

Since $p_{1}+p_{2}+p_{3}=8$, this is a contradiction.
Thus, we can only have $p_{3}=1$.
Moreover, $p_{1}$ cannot be less than or equal to 4, otherwise, five events would score at most 20 points, which contradicts $A$ scoring 22 points.
Thus, $p_{1} \geqslant 5$.
If $p_{1} \geqslant 6$, then $p_{2}+p_{3} \leqslant 2$, and since $p_{2} \neq p_{3}$, this is impossible.
So
$$
p_{2}+p_{3} \geqslant 2+1=3,
$$

Thus, we can only have $p_{1}=5$, at this point only $p_{2}=2, p_{3}=1$.
From this, we can see that $A$ must have won four first places and one second place, at this point
$$
4 p_{1}+p_{2}=4 \times 5+2=22.
$$
$B$ has one first place, to make $B$ score 9 points, then there are four third places, at this point $p_{1}+4 p_{3}=5+4 \times 1=9$.
Thus, $C$ has four second places and one third place, at this point $4 p_{2}+p_{3}=9$.
From this, we know that $A$ is second in the 100-meter dash, $C$ is third in the 100-meter dash, and in the other events, $C$ is second. Therefore, in the high jump, $C$ is second.