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

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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>
 
<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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==Solutions==
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==Solution 1==
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===Solution 1===
 
 
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>.
 
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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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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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>.
  
 
Next, <math>M=6</math> does not work because <math>52</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>.
  
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We try <math>M=7</math> <math>does</math> work by 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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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.
  
 
<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.
 
<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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===Solution 2===
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==Solution 2==
 
<math>76=3N+4M > 10M</math>, giving <math>M \le 7</math>.
 
<math>76=3N+4M > 10M</math>, giving <math>M \le 7</math>.
 
Since <math>M>4</math>, we have <math>M=5,6,7</math>.
 
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>.
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This gives <math>3N=48</math>, as desired. The answer is <math>\boxed{\textbf{(B)}~48}</math>
 
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===Video Solutions===
 
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https://youtu.be/LiAupwDF0EY - Happytwin
 
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https://www.youtube.com/watch?v=bJSWtw91SLs - Oliver Jiang
 
  
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https://youtu.be/HISL2-N5NVg?t=4968 - pi_is_3.14
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This gives <math>3N=48</math>, as desired. The answer is <math>\boxed{\textbf{(B)}~48}</math>.
  
 
==See Also==
 
==See Also==

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
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All AJHSME/AMC 8 Problems and Solutions

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