Difference between revisions of "2015 AMC 8 Problems/Problem 24"
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== Problem == | == Problem == | ||
− | A baseball league consists of two four-team divisions. Each team plays every other team in its division <math>N</math> games. Each team plays every team in the other division <math>M</math> games with <math>N>2M</math> and <math>M>4</math>. Each team plays | + | A baseball league consists of two four-team divisions. Each team plays every other team in its division <math>N</math> games. Each team plays every team in the other division <math>M</math> games with <math>N>2M</math> and <math>M>4</math>. Each team plays 76 games. How many games does a team play within its own division? |
<math>\textbf{(A) } 36 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 54 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 72 </math> | <math>\textbf{(A) } 36 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 54 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 72 </math> |
Revision as of 03:13, 14 January 2022
Problem
A baseball league consists of two four-team divisions. Each team plays every other team in its division games. Each team plays every team in the other division games with and . Each team plays 76 games. How many games does a team play within its own division?
Solutions
Solution 1
On one team they play games in their division and games in the other. This gives .
Since we start by trying . This doesn't work because is not divisible by .
Next, does not work because is not divisible by .
We try work by giving and thus games in their division.
seems to work, until we realize this gives , but so this will not work.
Solution 2
, giving . Since , we have . Since is , we must have equal to , so .
This gives , as desired. The answer is
Video Solutions
https://youtu.be/LiAupwDF0EY - Happytwin
https://www.youtube.com/watch?v=bJSWtw91SLs - Oliver Jiang
https://youtu.be/HISL2-N5NVg?t=4968 - pi_is_3.14
See Also
2015 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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.