Difference between revisions of "2012 AMC 10B Problems/Problem 15"
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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. | ||
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Revision as of 12:15, 4 July 2013
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.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
