Nováková, Vaňková, and Sudková won the relay and, in addition to diplomas, they also received a box of chocolates, which they immediately peeled after the race. If Petra had eaten 3 more chocolates, she would have eaten as many as Míša and Jana together. And if Jana had enjoyed 7 more chocolates, she would have had as many as the other two together. We also know that the number of chocolates Vaňková ate is divisible by three and that Sudková indulged in seven chocolates. What were the girls' names? How many chocolates did each of them eat?

( $M$. Volfová)
First method: Let the number of candies Petra ate be $p$, Jana $j$, and Míša $m$. We have:

$$
\begin{aligned}
p+3 & =m+j, \\
j+7 & =p+m .
\end{aligned}
$$

Subtract the second equation from the first. We get $p-j-4=j-p$, thus:

$$
2 p-4=2 j, \quad p=j+2 .
$$

We construct a table:

| $j$ | 1 | 2 | $\ldots$ | 5 | 6 | 7 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $p$ | 3 | 4 | $\ldots$ | 7 | 8 | 9 |
| $m$ | 5 | 5 | $\ldots$ | 5 | 5 | 5 |

The number 7 appears in only two columns; one number must be divisible by three — the last column fits. Jana Sudková had 7 candies; Petra Vaňková 9, and Míša Nováková 5 candies.

(finding the relationship $p=j+2 \ldots 2 b$., constructing the table ... 2 b., choosing the correct column ... 1 b., correct assignment of names ... 1 b.)

SECOND METHOD: In the same way, we arrive at the relationship $p=j+2$. We substitute this expression for $p$ into the first equation. After rearranging, we obtain $m$.

According to the problem, one of the unknowns is 7.

If $p=7$, then $j=p-2=5$. In this case, no unknown would be divisible by three.

If $j=7$, then $p=j+2=9$. This option satisfies the problem statement.

Jana Sudková had 7 candies; Petra Vaňková 9, and Míša Nováková 5 candies.

(finding the relationship $p=j+2 \ldots 2 b$.,

$m=5, j=7, p=9 \ldots 3$ b.,

correct assignment of names ... 1 b.)