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

(Solution)
Line 17: Line 17:
 
-----------------
 
-----------------
 
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.
 
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.
 +
 
==See Also==
 
==See Also==
  
 
{{AMC10 box|year=2012|ab=B|num-b=13|num-a=15}}
 
{{AMC10 box|year=2012|ab=B|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:22, 8 February 2014

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.

See Also

2012 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS