Difference between revisions of "2019 AMC 8 Problems/Problem 19"
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==Solution 3== | ==Solution 3== | ||
Revision as of 19:10, 17 October 2020
Solution 3
We can name the top three teams as and . We can see that , because these teams have the same points. If we look at the matches that involve the top three teams, we see that there are some duplicates: and come twice. In order to even out the scores and get the maximum score, we can say that in match and each win once out of the two games that they play. We can say the same thing for and . This tells us that each team and win and lose twice. This gives each team a total of 3 + 3 + 0 + 0 = 6 points. Now, we need to include the other three teams. We can label these teams as and . We can write down every match that or plays in that we haven't counted yet: and . We can say and win each of these in order to obtain the maximum score that and can have. If and win all six of their matches, and will have a score of . results in a maximum score of . This tells us that the correct answer choice is .
~Champion1234
Video Solutions
Associated Video - https://youtu.be/s0O3_uXZrOI
Video Solution - https://youtu.be/Lw8fSbX_8FU (Also explains problems 11-20)
See Also
2019 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.