Problem:
A baseball league consists of two four-team divisions. Each team plays every other team in its division N games. Each team plays every team in the other division M games with N>2M and M>4. Each team plays a 76 game schedule. How many games does a team play within its own division?
Answer Choices:
A. 36
B. 48
C. 54
D. 60
E. 72
Solution:
The number of games played by a team is 3N+4M=76. Because M>4 and N>2M it follows that N>8. Because 4 divides both 76 and 4M,4 must divide 3N and hence N. If N=12 then M=10 and the condition N>2M is not satisfied. If N≥20 then M≤4 and the condition M>4 is not satisfied. So the only possibility is N=16 and M=7. So each team plays 3⋅16=48 games within its division and 4⋅7=28 games against the other division.
OR
The total number of games played by each team is 3N+4M=76. Make a chart of possibilities with M>4:
M5678910111213​4M202428323640444852​76−4M=3N565248444036322824​N(not an integer)(not an integer)16(not an integer)(not an integer)12(not an integer)(not an integer)​​
Only M=7 and N=16 satisfy the conditions.
The case M=10 and N=12 violates the condition N>2M.
So each team plays 3N=48 divisional games and 4M=28 games against the other division.
This is modeled on the Pioneer Baseball League with teams in Colorado, Idaho, Montana, and Utah.
Answer: B​.
The problems on this page are the property of the MAA's American Mathematics Competitions
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