Difference between revisions of "2015 AMC 8 Problems/Problem 24"
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===Solution 2=== | ===Solution 2=== | ||
<math>76=3N+4M > 10M</math>, giving <math>M \le 7</math>. | <math>76=3N+4M > 10M</math>, giving <math>M \le 7</math>. | ||
− | Since <math>M>4</math>, we have <math>M=5,6,7</math> | + | Since <math>M>4</math>, we have <math>M=5,6,7</math>. |
Since <math>4M</math> is <math>1</math> <math>\pmod{3}</math>, we must have <math>M</math> equal to <math>1</math> <math>\pmod{3}</math>, so <math>M=7</math>. | Since <math>4M</math> is <math>1</math> <math>\pmod{3}</math>, we must have <math>M</math> equal to <math>1</math> <math>\pmod{3}</math>, so <math>M=7</math>. | ||
Revision as of 17:56, 22 October 2017
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 a 76 game schedule. How many games does a team play within its own division?
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 this does work giving and thus games in their division.
Solution 2
, giving . Since , we have . Since is , we must have equal to , so .
This gives , as desired. The answer is .
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 |
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