8. Alberto, Barbara, and Carlo are playing a game. This game consists of $k$ turns, at the end of each of which the first-place player receives $a_{1}$ points, the second-place player receives $a_{2}$ points, and the third-place player receives $a_{3}$ points, with $a_{1}>a_{2}>$ $a_{3}>0$ all being integers. The final scores of Alberto, Barbara, and Carlo are 22, 9, 9, respectively. Knowing that Barbara won the first turn, who came in second in the second turn?
(A) Necessarily Alberto
(B) Necessarily Barbara
(C) Necessarily Carlo
(D) Either Alberto or Barbara could have come in second
(E) Either Carlo or Barbara could have come in second
8. The answer is $\mathbf{( C )}$. We notice that during the game, a total of $k\left(a_{1}+a_{2}+a_{3}\right)$ points are assigned, so we can write $k\left(a_{1}+a_{2}+a_{3}\right)=22+9+9=40$, from which we deduce that $k$ and $a_{1}+a_{2}+a_{3}$ are both divisors of 40. We also notice that since $a_{1}, a_{2}$, $a_{3}$ are distinct positive integers and $a_{1}>a_{2}>a_{3}$, it must be that $a_{3} \geq 1, a_{2} \geq 2$ and $a_{1} \geq 3$, so $a_{1}+a_{2}+a_{3} \geq 6$. Therefore, we have $k \leq \frac{40}{6}$, and listing the divisors of $40$, $k$ can be $1,2,4,5$. Let's reason through the cases:

- $k=1$ is clearly impossible, otherwise we would have $a_{2}=a_{3}=9$, but by hypothesis $a_{2}>a_{3}$.
- $k=2$ is impossible, otherwise we would have $2 a_{1} \geq 22$ to give Alberto enough points, but also $a_{1}9$, which is absurd.
- $k=5$ is possible, indeed we have $a_{1}+a_{2}+a_{3}=8$ and the possible triples are $(4,3,1),(5,2,1)$. The first case is excluded because Alberto could not reach a score of 22 even with 5 wins. The second case, however, works, because to get 22, Alberto must have 4 wins and one second place, and since by hypothesis Barbara won the first round, the ranking of the first round must be Barbara-Alberto-Carlo. At this point, Carlo must have come in second in all the other rounds and therefore in particular in the second.