# 2015 AMC 8 Problems/Problem 24

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## 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?

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

## Solution 1

On one team they play $3N$ games in their division and $4M$ 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 $M=7$ does work by giving $N=16$ ,$~M=7$ and thus $3\times 16=\boxed{\textbf{(B)}~48}$ games in their division.

$M=10$ seems to work, until we realize this gives $N=12$, but $N>2M$ so this will not work.

## Solution 2

$76=3N+4M > 10M$, giving $M \le 7$. Since $M>4$, we have $M=5,6,7$. Testing all of the cases we get tha $M=7$.

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