# Difference between revisions of "2019 AMC 8 Problems/Problem 19"

## Solution 3

We can name the top three teams as $A, B,$ and $C$. We can see that $A=B=C$, 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: $AB, BC,$ and $AC$ come twice. In order to even out the scores and get the maximum score, we can say that in match $AB, A$ and $B$ each win once out of the two games that they play. We can say the same thing for $AC$ and $BC$. This tells us that each team $A, B,$ and $C$ 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 $D, E,$ and $F$. We can write down every match that $A, B,$ or $C$ plays in that we haven't counted yet: $AD, AD, AE, AE, AF, AF, BD, BD, BE, BE, BF, BF, CD, CD, CE, CE, CF,$ and $CF$. We can say $A, B,$ and $C$ win each of these in order to obtain the maximum score that $A, B,$ and $C$ can have. If $A, B,$ and $C$ win all six of their matches, $A, B,$ and $C$ will have a score of $18$. $18 + 6$ results in a maximum score of $\boxed{24}$. This tells us that the correct answer choice is $\boxed{C}$.

~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 byProblem 18 Followed byProblem 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

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