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

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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.
 
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>.
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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{\textbf{(B) }48}</math>.
  
 
===Solution 2===
 
===Solution 2===

Revision as of 12:28, 26 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 $3N=\boxed{\textbf{(B) }48}$.

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.

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 $\boxed{\textbf{(B)}~48}$.

See Also

2015 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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