2015 AMC 8 Problems/Problem 24

Revision as of 17:00, 25 November 2015 by Ariedel (talk | contribs)

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?

$\textbf{(A) } 36 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 54 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 72$

Solution 1

Note that the equation rewrites to $3N+4M=76$.

Now remark that if $(m,n)$ is a solution to this equation, then so is $(m+3,n-4)$. This is because \[3(n-4)+4(m+3)=3n-12+4m+12=3n+4m=76.\] Thus, we can now take an "edge case" solution and work upward until both conditions ($N>2M$ and $M>4$) are met.

We see by inspection that $(M,N)=(1,24)$ is a solution. By the above work, we can easily deduce that $(4,20)$ and $(7,16)$ are solutions. The last one is the intended answer (the next solution fails $N>2M$) so our answer is $3N=\boxed{48\textbf{ (B)}}$.

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