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

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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>.
 
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>.
  
−
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>.
+
Since <math>M>4</math> we start by trying <math>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>.
  
−
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.
+
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.
  
−
<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.
  
 
===Solution 2===
 
===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>.
+
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>.
  
−
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}$.
  
 
==Solution 3==
 
==Solution 3==

Revision as of 18:22, 9 December 2023

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?

$\textbf{(A) } 36 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 54 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 72$

Solutions

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,$ (Error compiling LaTeX. Unknown error_msg)M=6$does not work because$52$is not divisible by$3$.

We try$ (Error compiling LaTeX. Unknown error_msg)M=7$$ (Error compiling LaTeX. Unknown error_msg)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===$ (Error compiling LaTeX. Unknown error_msg)76=3N+4M > 10M$, giving$M \le 7$. Since$M>4$, we have$M=5,6,7$. Since$4M$is$1$$ (Error compiling LaTeX. Unknown error_msg)\pmod{3}$, we must have$M$equal to$1$$ (Error compiling LaTeX. Unknown error_msg)\pmod{3}$, so$M=7$.

This gives$ (Error compiling LaTeX. Unknown error_msg)3N=48$, as desired. The answer is$\boxed{\textbf{(B)}~48}$.

Solution 3

Notice that each team plays $N$ games against each of the three other teams in its division. So that's $3N$.

Since each team plays $M$ games against each of the four other teams in the other division, that's $4M$.

So $3N+4M=76$, with $M>4, N>2M$.

Let's start by solving this Diophantine equation. In other words, $N=\frac{76-4M}{3}$.

So $76-4M\equiv0 \pmod{3}$ (remember: $M$ must be divisible by 3 for $N$ to be an integer!). Therefore, after reducing $76$ to $1$ and $-4M$ to $2M$ (we are doing things in $\pmod{3}$), we find that $M\equiv1 \pmod{3}$.

Since $M>4$, so the minimum possible value of $M$ is $7$. However, remember that $N>2M$! To find the greatest possible value of M, we assume that $N=2M$ and that is the upper limit of $M$ (excluding that value because $N>2M$). Plugging $N=2M$ in, $10M=76$. So $M<7.6$. Since you can't have $7.6$ games, we know that we can only check $M=7$ since we know that since $M>4, M<7.6, M\equiv1 \pmod{3}$. After checking $M=7$, we find that it works.

So $M=7, N=16$. So each team plays 16 games against each team in its division. Since they are asking for games in it division, which equals $3n = 48$. Select $\boxed{\textbf{(B)}~48}$.

This might be too complicated. But you should know what's happening if you read the The Art of Problem Solving: Introduction to Number Theory by Mathew Crawford. Notice how I used chapter 12's ideas of basic modular arithmetic operations and chapter 14's ideas of solving linear congruences. Remember: the Introduction Series books by AoPS are for 6th-10th graders! So make sure to read the curriculum books!

This goes same for high school students as well, and especially for those who want to continue on their path of AIME/USA(J)MO.

~hastapasta

Video Solution by OmegaLearn

https://youtu.be/HISL2-N5NVg?t=4968

Video Solutions

https://youtu.be/LiAupwDF0EY - Happytwin

https://www.youtube.com/watch?v=bJSWtw91SLs - Oliver Jiang

https://youtu.be/0ZMDsuOYGqE

~savannahsolver

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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