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

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\textbf{(E) } 72
 
\textbf{(E) } 72
 
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===Solution 1===
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Note that the equation rewrites to <math>3N+4M=76</math>.
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Now remark that if <math>(m,n)</math> is a solution to this equation, then so is <math>(m+3,n-4)</math>.  This is because <cmath>3(n-4)+4(m+3)=3n-12+4m+12=3n+4m=76.</cmath> Thus, we can now take an "edge case" solution and work upward until both conditions (<math>N>2M</math> and <math>M>4</math>) are met.
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We see by inspection that <math>(M,N)=(1,24)</math> is a solution.  By the above work, we can easily deduce that <math>(4,20)</math> and <math>(7,16)</math> are solutions.  The last one is the intended answer (the next solution fails <math>N>2M</math>) so our answer is <math>3N=\boxed{48\textbf{ (B)}}</math>.

Revision as of 17:00, 25 November 2015

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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