Difference between revisions of "2015 AMC 8 Problems/Problem 24"
(→Solution 2) |
Crocodile 40 (talk | contribs) (→Solution 2) |
||
Line 17: | Line 17: | ||
===Solution 2=== | ===Solution 2=== | ||
− | On one team they play <math>\binom{3}{2}N</math> games in their | + | On one team they play <math>\binom{3}{2}N</math> games in their divisionnnnnn and <math>4(M)</math> games in the other. This gives <math>3N+4M=76</math> |
Since <math>M>4</math> we start by trying <math>M=5</math>. This doesn't work because <math>56</math> is not divisible by <math>3</math>. | Since <math>M>4</math> we start by trying <math>M=5</math>. This doesn't work because <math>56</math> is not divisible by <math>3</math>. |
Revision as of 00:01, 10 June 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?
Contents
Solution 1
Note that the equation rewrites to .
Now remark that if is a solution to this equation, then so is . This is because Thus, we can now take an "edge case" solution and work upward until both conditions ( and ) are met.
We see by inspection that is a solution. By the above work, we can easily deduce that and are solutions. The last one is the intended answer (the next solution fails ) so our answer is .
Solution 2
On one team they play games in their divisionnnnnn 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 3
, 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.