In a one-off football championship, four teams participated: $A, B, C$, and $D$. The championship is not yet over. Based on the matches played so far, the following incomplete table was created, where the different letters represent distinct integers between 0 and 9.

|  | PLAYED | WON | DRAWN | LOST | SCORED | CONCEDED | POINTS |  |  |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
|  | MATCHES |  |  |  |  |  |  | GOALS |  |  |
| $A$ | $\mathrm{P}$ | $\mathrm{S}$ | $\mathrm{K}$ | $\mathrm{H}$ | $\mathrm{T}$ |  | $\mathrm{M}$ |  |  |
| $B$ |  | $\mathrm{H}$ |  |  | $\mathrm{M}$ | $\mathrm{M}$ |  |  |  |
| $C$ | $\mathrm{~S}$ |  | $\mathrm{~K}$ |  | $\mathrm{H}$ | $\mathrm{P}$ |  |  |  |
| $D$ | $\mathrm{~K}$ |  |  |  |  |  |  |  |  |

What matches have been played in the championship so far, and what were the results? (2 points for a win, 1 point for a draw, and 0 points for a loss.)
P can be expressed as the sum of three different non-negative integers $(S, K, H)$, so its value is at least 3. But it cannot be more, since the tournament is a round-robin, and there are four teams; thus, $P=3$. This means that the other teams have already played against $A$, hence according to the first column, $S>0, K>0$. On the other hand, it is clear that $S>K$, as equality is not possible: thus, team $C$ must have played more matches than the number of these that ended in a draw. Given these, it is uniquely determined that $S=2, K=1, H=0$, and from this, team $A$'s points $M=5$.

Team $C$ did not score any goals $(H=0)$, but received three. One of their matches ended in a draw $(K=1)$, which could not have been anything other than 0:0. Therefore, the other match must have ended 0:3. Since $B$ did not win and $D$ only played against $A$, $C$ must have received the 0:3 loss from $A$, and the draw was against $B$.

Team $B$, with a goal difference of 5:5, could not have lost, as they had no wins - only draws. One of these was against $C$, and they did not play against $D$, so the 5:5 draw must have been against $A$.

Finally, let's look at the $D-A$ match! $A$ won this, because $A$ played their draw against $B$ and did not lose. So far, we have accounted for 8 goals, so $D$ could have scored only one more, due to the constraint on the values of the letters. Thus, $T=9$.

Answer to the question of the problem: four matches were played, with the following results:

$$
\begin{array}{llll}
A-B & 5: 5 & A-D & 1: 0 \\
A-C & 3: 5 & B-C & 0: 0
\end{array}
$$

Remarks. 1. Many correctly noted that the problem is only solvable if the "numbers between 0 and 9" include 0 and 9 as well.

2. Some arrived at the correct solution based on "non-mathematical" grounds, for example, by using the fact that in tournament match results, teams are listed in order of their standings. This information was not provided in the problem statement, so they might have missed another solution. Moreover, the problem is solvable without this. Those who heavily relied on this condition in multiple places did not receive the maximum score.