Difference between revisions of "2015 AMC 8 Problems/Problem 24"
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We try <math>M=7</math> this does work giving <math>N=16,~M=7</math> and thus <math>3\times 16=\boxed{\textbf{(B)}~48}</math> games in their division. | We try <math>M=7</math> this does work giving <math>N=16,~M=7</math> and thus <math>3\times 16=\boxed{\textbf{(B)}~48}</math> games in their division. | ||
| − | ===Solution | + | ===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> | ||
Revision as of 14:54, 4 July 2016
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
