Blinkov A.d:

a) In a football tournament in a round-robin format, 75 teams participated. A team received 3 points for a win, 1 point for a draw, and 0 points for a loss. It is known that every two teams scored a different number of points. Find the smallest possible difference in points between the teams that took first and last place.

b) The same question for $n$ teams.
b) The minimum gap between the first and last place cannot be less than $n-1$. We will prove that for $n>3$ it is possible to construct a tournament scheme to achieve a gap of $n-1$ (if $n=2$ or 3, the minimum gap is obviously 3 points).

By induction, we will construct a table for $n$ teams, where the results are all integers from $n-2$ to $2n-3$. Base case ($n=4$). Example of a tournament table for four teams:

| Team | 1 | 2 | 3 | 4 | Points |
| :--- | :---: | :---: | :---: | :--- | :--- |
| 1 | $\times$ | 3 | 1 | 1 | 5 |
| 2 | 0 | $\times$ | 1 | 3 | 4 |
| 3 | 1 | 1 | $\times$ | 1 | 3 |
| 4 | 1 | 0 | 1 | $\times$ | 2 |

Inductive step. By the inductive hypothesis, there is such a table for $n$ teams. We will divide the teams into triples based on the number of points they have: the first triple consists of teams with $2n-3, 2n-4, 2n-5$ points, the second triple consists of teams with $2n-6, 2n-7, 2n-8$ points, and so on (the last triple may be incomplete). Add one more team. Consider three cases.

![](https://cdn.mathpix.com/cropped/2024_05_06_ed128c2d5b4d71db92ebg-11.jpg?height=58&width=1935&top_left_y=1506&top_left_x=39)

Then the first team will have

$2n-3$ points, the second team will have $2n-1$ points and will move to the first place. The third team will have $2n-2$ points. Similarly, we will proceed with the other complete triples. Then all triples will shift "up" by two points. There will be one team left that was in the last place with $n-2$ points, and the new team will have $n-1$ points. Let them play a draw against each other and add one point each.

2) $n=3k+2$. Teams from $k$ triples play against the new team similarly to case 1). The last two teams had $n-1$ and $n-2$ points, and the new team has $n-2$ points. Let the last team win against the new team, and the second-to-last team play a draw with the new team:
3) $n=3k$. After the new team plays against the first $k-1$ triples, it will have $n-3$ points. The last three teams had $n, n-1$,

$n-2$ points. Let the second-to-last team win against the new team, the last team play a draw with the new team, and the team with $n$ points lose. Then the former last teams will have $n, n+2, n-1$ points, and the new team will have $n+1$ points.

## Answer

a) 74; b) $n-1$, if $n>3$; 3, if $n=2$ or 3.