$36 \cdot 57$ Distribute the four cards $2, 4, 7, K$ to four people, each person scores according to the number on the card ($K$ is counted as 13), then the cards are collected, reshuffled, and distributed and scored again, $\cdots$, after several rounds, it is found that the four people have accumulated scores of $16, 17, 21$ and 24. It is known that the person who scored 16 points last scored 2 points. Then his score in the first round is
(A) 2 .
(B) 4 .
(C) 7
(D) 13 .
(2nd "Five Sheep Cup" Junior High School Mathematics Competition, 1990)
[Solution] From the problem, we know that the total score of the four people each time is
$$
2+4+7+13=26 \text{, }
$$

After several rounds, the cumulative scores of the four people are
$$
16+17+21+24=78 \text{. }
$$

It can be seen that the cards were dealt three times $(78 \div 26=3)$. The person who scored 16 points last scored 2 points in the last round, so the total score of the first two rounds is 14 points.
Among $2,4,7,13$, only two 7s can sum up to 14.
Therefore, the answer is (C).