# Difference between revisions of "2012 AMC 10B Problems/Problem 15"

## Problem

In a round-robin tournament with 6 teams, each team plays one game against each other team, and each game results in one team winning and one team losing. At the end of the tournament, the teams are ranked by the number of games won. What is the maximum number of teams that could be tied for the most wins at the end on the tournament?

$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6$

## Solution

The total amount of games in the tournament is 5+4+3+2+1=15. Now, we see which numbers from 1-6 divide 15, and it seems 1,3, and 5 do. 5 is the largest number, so (D) 5 is the correct answer. Here's a poorly done chart of 5 games won:

_ _ _ _ _ _ _


| 1 2 3 4 5 6 | |1X WL WL W | |2 L XW L WW| |3 WL X WL W| |4 LW LX W W| |5 WL W LX W| |6 L L L L L X| The "x's" are for when it is where a team is versus itself, which cannot happen. The chart says that Team 6 has lost all of it's matches, which means that each team gets one win. Then, alternating Wins and Losses were tried. It shows that it is possible for 5 teams to tie for the same amount of wins, which in this case is 2 wins.