Several teams are competing in a basketball tournament, and each team plays every other team once. Each game has exactly 1 winner and 1 loser (no ties). If 4 teams lost exactly 5 games, 5 teams won exactly 3 games, and each of the remaining teams won all of its games, what is the total number of games played during the tournament?

A) 36
B) 45
C) 55
D) 66
E) 78

*kudos for all correct solutions
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Hi..
Brent, shifted it to PS
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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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