# Difference between revisions of "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?

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

## Contents

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

### Solution 2

On one team they play $\binom{3}{2}N$ games in their divisionnnnnn 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 $\boxed{\textbf{(B)}~48}$.