Several teams are competing in a basketball tournament, and each team

Another way..
Number of matches lost = number of matches won

Number of matches lost
1) 4 teams lost 5 matches so 4*5=20
2) 5 teams won 3 so lost (n-1-(3))=n-4, thus 5*(n-4)
3) all remaining teams won all matches. So 0
Total matches lost = 20+5*(n-4)=20+5n-20=5n

Number of matches won
1) 4 teams lost 5 matches so they won (n-1-(5))=n-6, thus 4*(n-6)
2) 5 teams won 3 so 5*3=15
3) all remaining teams won all matches. Each team plays n-1 matches a there are n teams. Now remaining teams will be n-(4+5)=n-9.. so (n-9)(n-1)
Total matches won = 4*(n-6)+15+(n-9)(n-1)=4n-24+15+n^2-10n+9=n^2-6n

Thus won=lost....5n=n^2-6n......n^2-11n=0....n(n-11)=0
So n 0 or 11, but there are more than 9 teams so n=11

Matches =11C2=11*10/2=55

C

I have changed the question stem slightly, because in the original wording it would be impossible to each of two teams win all its games (one played the other)!

Very nice problem, Brent. Congrats! (kudos!)

Let´s suppose there are (N+1) teams. Hence each one will play the N other teams, so that

\(? = {{\left( {N + 1} \right) \cdot N} \over 2}\)

Each of 5 teams lost 5 games and won (N-5) games.
Each of 4 teams lost (N-3) games and won 3 games.
Each of the remaining (N+1-9) teams lost 1 game and won N-1 games. (*)

(*) If there were at least one SINGLE team who lost exactly 5 games AND won exactly 3 games, we would have N=8 (hence 9 teams), impossible.
(No team would have lost exactly one game, contradicting the question stem.)

There were no ties, therefore the number of games lost must equal the number of games won, hence:

\(5 \cdot 5 + 4\left( {N - 3} \right) + \left[ {\left( {N + 1} \right) - 9} \right] \cdot 1 = 5\left( {N - 5} \right) + 4 \cdot 3 + \left[ {\left( {N + 1} \right) - 9} \right] \cdot \left( {N - 1} \right)\)

\({N^2} - 9N - 10 = 0\,\,\,\,\mathop \Rightarrow \limits^{{\rm{sum}} = 9\,\,,\,\,\,{\rm{product}} = - 10} \,\,\,\,\,N = 10\,\,\,\,{\rm{or}}\,\,\,\,N = - 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,N = 10\)

\(? = {{\left( {N + 1} \right) \cdot N} \over 2} = 55\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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