C14 (16-1, USA) $A, B, C$ are playing a game: On 3 cards, integers $p, q, r (0<p<q<r)$ are written. The 3 cards are shuffled and distributed to $A, B, C$, each getting one. Then, according to the number on each person's card, they are given marbles. After that, the cards are collected, but the marbles are kept by each person. This process is repeated for more than two rounds (each round includes shuffling the cards, distributing the cards, giving marbles, and collecting the cards). After the final round, $A, B, C$ have a total of 20, 10, 9 marbles, respectively. It is known that $B$ received $r$ marbles in the last round. Who received $q$ marbles in the first round?
Let the game have proceeded for $n$ rounds, then
$$
n(p+q+r)=20+10+9=39=3 \times 13,
$$

By the problem statement, $n \geqslant 2, p+q+r \geqslant 1+2+3=6$, so $n=3, p+q+r=13$.
Since $r \leqslant 10-1-1=8,3 r>20$, i.e., $r \geqslant 7$, thus $r=7$ or 8.
If $r=7$, then $p+q=6$, and the sum of the marbles obtained in the first two rounds is 3, so $2 p \leqslant 3, p <2$, hence $p=1, q=5$, in this case, $B$ cannot obtain a sum of 3 in the first two rounds.

If $r=8$, then $p+q=5$, and the sum of the marbles obtained by $B$ in the first two rounds is 2, so $2 p \leqslant 2$, thus $p=1, q=4$.
For $A$, we have
$$
20 \leqslant 2 r+q=20,
$$

Therefore, the marbles obtained by $A$ in the second round are $8,8,4$, and the marbles obtained in the first and second rounds are 8, and the marbles obtained in the third round are 4, thus $C$ obtains 4 marbles in the first round.