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Difference between revisions of "2015 AMC 8 Problems/Problem 24"

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

Solutions

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 $M=5,6,7$. 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 $\boxed{\textbf{(B)}~48}$

Video Solutions

https://youtu.be/LiAupwDF0EY - Happytwin

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