Difference between revisions of "2019 AMC 8 Problems/Problem 19"
(→Problem 19) |
A1337h4x0r (talk | contribs) (→Solution 1) |
||
| Line 6: | Line 6: | ||
==Solution 1== | ==Solution 1== | ||
| + | After fully understanding the problem, we immediately know that the three top teams, say team A, team B, and team C, must beat the other three teams D,E,F. Therefore, A,B,C must each obtain <math>(3+3+3)=9</math> points. However, they play against each team twice, for a total of <math>18</math> points against D,E and,F. | ||
==See Also== | ==See Also== | ||
Revision as of 19:01, 20 November 2019
Problem 19
In a tournament there are six team that play each other twice. A team earns 
points for a win, 
point for a draw, and 
points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for each of the top three teams?
Solution 1
After fully understanding the problem, we immediately know that the three top teams, say team A, team B, and team C, must beat the other three teams D,E,F. Therefore, A,B,C must each obtain 
points. However, they play against each team twice, for a total of 
points against D,E and,F.
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. 
