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Difference between revisions of "2015 AMC 8 Problems/Problem 24"
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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>.
 
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>.
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 +
===Solution 2===
 +
On one team they play <math>\binom{3}{2}N</math> games in their division and <math>4(M)</math> games in the other.  This gives <math>3N+4M=76</math>
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 +
Since <math>M>4</math> we start by trying <math>M=5</math>. This doesn't work because <math>56</math> is not divisible by <math>3</math>.
 +
 +
Next <math>M=6</math>, does not work because <math>52</math> is not divisible by <math>3</math>
 +
 +
We try <math>M=7</math> this does work giving <math>N=16,~M=7</math> and thus <math>3\times 16=\boxed{\textbf{B},~48}</math> games in their division.

Revision as of 16:03, 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 $3N=\boxed{48\textbf{ (B)}}$.

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.

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