5. Nika and Tim were playing cards. An undecided outcome was not possible. They had agreed in advance that the winner of each game would receive more points than the loser and that the loser would receive a positive number of points. They determined how many points the winner would receive after each game and how many the loser would receive. After several games, Nika had 30 points, and Tim, who won 2 games, had 25 points. How many points did the winner receive after each game?

The tasks should be solved independently. You have 150 minutes for solving. The use of notes, literature, or a pocket calculator is not allowed.

Mathematical Competition for High School Students in Slovenia

Optional Competition

March 31, 2004

## Tasks for 2nd Year
I/5. After a few games, Nika and Tim together have $30+25=55$ points. Let's say the winner of each game gets $m$ points, and the loser gets $n$ points. Since after each game they gain a total of $m+n$ points and they have played more than 2 games, $m+n=11$ or $m+n=5$.

If $m+n=5$, Nika and Tim have played 11 games, of which Tim won 2. Since $m>n>0$, there are only 2 possibilities, namely $m=4, n=1$ or $m=3, n=2$. In the first case, Tim would have $2 \cdot 4+9 \cdot 1=17$ points, and in the second case, $2 \cdot 3+9 \cdot 2=24$ points, but neither of these point totals matches the data from the problem.

If $m+n=11$, Nika and Tim have played 5 games, of which Tim won 2. In this case, there are several possibilities for the pair $(m, n)$, namely $(10,1),(9,2),(8,3),(7,4)$, and $(6,5)$. Since $2 m+3 n=25$, we find that only the pair $(8,3)$ satisfies the condition. Therefore, the winner of each game gets $m=8$ points.

Conditions $m+n=11, m+n=5$ or equivalent: 2 points. Analysis of the case $m+n=5$: 2 points. Analysis of the case $m+n=11: 2$ points. Solution: $m=8: 1$ point.

If the competitor only states the solution $m=8$, but it is not clear from the process that this is the only solution: 1 point. If the competitor correctly analyzes all possible cases, they can also analyze all possibilities $1 \leq m<\frac{25}{2}: 7$ points.