2015 AMC 8 Problems/Problem 24

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 $3N=\boxed{48\textbf{ (B)}}$ .

Solution 2

On one team they play $\binom{3}{2}N$ games in their division and $4(M)$ games in the other. This gives $3N+4M=76$

Since $M>4$ we start by trying $M=5$ . This doesn't work because $56$ is not divisible by $3$ .

Next $M=6$ , does not work because $52$ is not divisible by $3$

We try $M=7$ this does work giving $N=16,~M=7$ and thus $3\times 16=\boxed{\textbf{(B)},~48}$ games in their division.

Solution 3

$76=3N+4M > 10M$ , giving $M \le 7$ . Since $M>4$ , we have $M=5,6,7$ Since $4M$ is $1$ $\pmod{3}$ , we must have $M$ equal to $1$ $\pmod{3}$ , so $M=7$ .

This gives $3N=48$ , as desired. The answer is $\boxed{\textbf{(B)}~48}$ .

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2015 AMC 8 Problems/Problem 24

Solution 1

Solution 2

Solution 3