12. Nick and Tim are playing a game, in which there are no ties in each round. In a round of the game, the losing side must take at least one candy, and the winning side takes more candies than the losing side. They also agreed in advance on the number of candies (two given positive integers) that the losing and winning sides should take in each round. After several rounds, Nick has a total of 30 candies, and Tim has a total of 25 candies. It is known that Tim won 2 rounds. Question: How many candies should the winning side take in each round?
(2004 Slovenian Olympiad Problem)
12. After several rounds of the game, the two people have a total of $30+25=55$ candies. Suppose in each round, the winner takes $m$ candies, and the loser takes $n$ candies, with the two of them taking a total of $m+n$ candies. When the game lasts more than 2 rounds, there are two possibilities: $m+n=5$ or $m+n=11$. If $m+n=5$, then Nick and Tim played a total of 11 rounds, with Tim winning 2 rounds. Since $m>n>0$, there are 2 possibilities: $m=4, n=1$ or $m=3, n=2$. In the first scenario, Tim would take $2 \times 4 + 9 \times 1 = 17$ candies, and in the second scenario, Tim would take $2 \times 3 + 9 \times 2 = 24$ candies, both of which contradict the given data. If $m+n=11$, then Tim and Nick played a total of 5 rounds, with Tim winning 2 rounds. The pairs $(m, n)$ have the following 5 possibilities: $(10,1),(9,2),(8,3),(7,4),(6,5)$. Since $2m+3n=25$, only the pair $(8,3)$ satisfies this requirement. Therefore, in each round, the winner should take $m=8$ candies. (Note: This can also be derived by solving the system of equations $\left\{\begin{array}{l}m+n=11 \\ 2m+5m=25\end{array}\right.$, yielding $\left\{\begin{array}{l}m=8 \\ n=3\end{array}\right)$.)