André, Bruno, Celina, and Dalva won a total of 21 medals in a competition. André won the most medals, Bruno won twice as many as Celina, and Dalva won three more than Bruno. How many medals could each of them have won?
Let $A, B, C$, and $D$ denote the number of medals won by André, Bruno, Celina, and Dalva, respectively. Then $A, B, C$, and $D$ are non-negative integers and $A+B+C+D=21$. We have that

- Bruno won twice as many medals as Celina, i.e., $B=2 C$.
- Dalva won three more medals than Bruno, i.e., $D=B+3=2 C+3$.

Thus, by assigning any value to $C$, we automatically know the values of $B$ and $D$. But $A+2 C+C+2 C+3=A+B+C+D=21$. Therefore, $A+5 C=18$ and, consequently, we can express $A$ in terms of $C$ as $A=18-5 C$. Note that $C \leq 3$, because if $C=4$, then $A=18-20=-2$, which is impossible.

Since $C$ is an integer greater than or equal to 0 and less than 4, we have only the following possibilities.

| $C$ | $A$ | $B$ | $D$ |
| :---: | :---: | :---: | :---: |
| 0 | 18 | 0 | 3 |
| 1 | 13 | 2 | 5 |
| 2 | 8 | 4 | 7 |
| 3 | 3 | 6 | 9 |

Since André was the one who received the most medals, $C=3$ does not work. The problem, therefore, has three possible solutions, listed below.

| $A$ | $B$ | $C$ | $D$ |
| :---: | :---: | :---: | :---: |
| 18 | 0 | 0 | 3 |
| 13 | 2 | 1 | 5 |
| 8 | 4 | 2 | 7 |