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

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A baseball league consists of two four-team divisions. Each team plays every other team in its division <math>N</math> games. Each team plays every team in the other division <math>M</math> games with <math>N>2M</math> and <math>M>4</math>. Each team plays a 76 game schedule. How many games does a team play within its own division?
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== Problem ==
  
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<math>
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A baseball league consists of two four-team divisions. Each team plays every other team in its division <math>N</math> games. Each team plays every team in the other division <math>M</math> games with <math>N>2M</math> and <math>M>4</math>. Each team plays a <math>76</math> game schedule. How many games does a team play within its own division?
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\textbf{(A) } 36 \qquad
 
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\textbf{(B) } 48 \qquad
 
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\textbf{(C) } 54 \qquad
 
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\textbf{(D) } 60 \qquad
 
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\textbf{(E) } 72
 
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</math>
 
  
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===Solution 1===
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<math>\textbf{(A) } 36 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 54 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 72 </math>
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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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==Solution 1==
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On one team they play <math>3N</math> games in their division and <math>4M</math> games in the other.  This gives <math>3N+4M=76</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>.
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Since <math>M>4</math> we start by trying M=5. This doesn't work because <math>56</math> is not divisible by <math>3</math>.
  
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===Solution 2===
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Next, <math>M=6</math> does not work because <math>52</math> is not divisible by <math>3</math>.
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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>.
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We try <math>M=7</math> does work by giving <math>N=16</math> ,<math>~M=7</math> and thus <math>3\times 16=\boxed{\textbf{(B)}~48}</math> games in their division.
  
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Next <math>M=6</math>, does not work because <math>52</math> is not divisible by <math>3</math>
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<math>M=10</math> seems to work, until we realize this gives <math>N=12</math>, but <math>N>2M</math> so this will not work.
  
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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.
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==Solution 2==
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===Solution 3===
 
 
<math>76=3N+4M > 10M</math>, giving <math>M \le 7</math>.
 
<math>76=3N+4M > 10M</math>, giving <math>M \le 7</math>.
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Since <math>M>4</math>, we have <math>M=5,6,7</math>
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Since <math>M>4</math>, we have <math>M=5,6,7</math>.
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Since <math>4M</math> is <math>1</math> <math>\pmod{3}</math>, we must have <math>M</math> equal to <math>1</math> <math>\pmod{3}</math>, so <math>M=7</math>.
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Testing all of the cases we get tha <math>M=7</math>.
  
 
This gives <math>3N=48</math>, as desired. The answer is <math>\boxed{\textbf{(B)}~48}</math>.
 
This gives <math>3N=48</math>, as desired. The answer is <math>\boxed{\textbf{(B)}~48}</math>.

Latest revision as of 09:01, 23 July 2024

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

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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