Just to confirm, does this mean that the points form an arithmetic progression.

achen29 wrote: Just to confirm, does this mean that the points form an arithmetic progression. i thinks so, Any solution?

AlastorMoody wrote: In a soccer tournament each team plays exactly one game with all others. The winner gets $3$ points, the loser gets $0$ and each team gets $1$ point in case of a draw. It is known that $n$ teams ($n \geq 3$) participated in the tournament and the final classification is given by the arithmetical progression of the points, the last team having only 1 point. Prove that this configuration is unattainable when $n=12$ Find all values of $n$ and all configurations when this is possible Interesting combi here is my solution for storage (assuming that I have understood the problem correctly). We will immediately solve the general case which will obviously imply part (a). Let $A_1,..,A_n$ be the participants of the tournament, let $a_i$ denote the number of points scored by participant $A_i$ such that $a_1 > a_2 >...>a_n$. Let $S$ be the total amount of points in the tournament, i.e, $S = a_1 + a_2 + .... + a_n$. Let $d \in \mathbb{N}$ be the common difference in the sequence. Let $a$ be the number of decisive games in the tournament (games with a winner) and let $b$ be the number of draws in the tournament. Claim: $d \neq 2$ is not possible Proof) Assume, $d=1$, we have that $a_i = n-i+1$ for all $i \in \{1,...,n\}$ meaning that $$S = \sum_{i=1}^{n} (n-i+1) = \frac{n(n+1)}{2}$$We also know that $S \geq 2 {n \choose 2}$ because we know that each match between any two teams must contribute at least 2 points to the sum of all points. Then $\frac{n(n+1)}{2} \geq n(n-1) \implies n \leq 3$ after dividing by $n$ and rearranging. Assume $n \neq 3$ which we will cover later. Now we can move on to the case where $d \geq 3$. FTSOC assume $d\geq 3$. Now we know that $a_1 = (n-1)d + 1 = nd - d +1 \geq 3n-2$ for all $d \geq 3, n \geq 1$. Moreover, we know that $a_1 \leq 3(n-1) = 3n -3 < 3n-2 \leq a_1 $ as a person can get at most $3$ points in each game which is clearly a contradiction and therefore $d \neq 3$. Claim: $n \leq 6$ Proof) We shall now look at the case where $d=2$. We know that $3a+2b = S$ by definition and that $a+b = {n \choose 2}$. As $d=2$, we know that $a_i = 2n - 2i +1$ and therefore $$S = \sum_{i=1}^{n} (2n-2i+1) = \sum_{j=1}^{n} 2j-1 = \sum_{j=1}^{2n} j - 2\sum_{j=1}^{n} j = \frac{2n(2n+1)}{2} - 2\frac{n(n+1)}{2} = n^2$$Then $3a + 2b = n^2$ and $a+b = \frac{n(n-1)}{2}$ and therefore $a = n$. Now, looking at $A_n$ we can see that $a_n = 1$ means $A_n$ must have drawn 1 game and lost $n-2$ games and therefore there are $2$ more decisive games among $A_1,..,A_{n-1}$. Now, simply looking at $A_{n-1}$ we can see that $a_{n-1} = 3$ means that $A_{n-1}$ must have lost at least $n-4$ and so, $n-4 \leq 2$, so $n \leq 6$ (this implies part (a)). Now, one can simply check the cases $n=3,4,5,6$ and see that $n=3,5,6$ do not work. There is a simple construction for $n=4$ $\square$