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

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==Solution 3==
 
 
 
We can name the top three teams as <math>A, B,</math> and <math>C</math>. We can see that <math>A=B=C</math>, 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: <math>AB, BC,</math> and <math>AC</math> come twice. In order to even out the scores and get the maximum score, we can say that in match <math>AB, A</math> and <math>B</math> each win once out of the two games that they play. We can say the same thing for <math>AC</math> and <math>BC</math>. This tells us that each team <math>A, B,</math> and <math>C</math> 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 <math>D, E,</math> and <math>F</math>. We can write down every match that <math>A, B,</math> or <math>C</math> plays in that we haven't counted yet: <math>AD, AD, AE, AE, AF, AF, BD, BD, BE, BE, BF, BF, CD, CD, CE, CE, CF,</math> and <math>CF</math>. We can say <math>A, B,</math> and <math>C</math> win each of these in order to obtain the maximum score that <math>A, B,</math> and <math>C</math> can have. If <math>A, B,</math> and <math>C</math> win all six of their matches, <math>A, B,</math> and <math>C</math> will have a score of <math>18</math>. <math>18 + 6</math> results in a maximum score of <math>\boxed{24}</math>. This tells us that the correct answer choice is <math>\boxed{C}</math>.
 
 
~Champion1234
 
 
 
 
== Video Solutions ==
 
== Video Solutions ==
  

Revision as of 19:11, 17 October 2020

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 (ProblemsAnswer KeyResources)
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

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