Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Problem)
Line 23: Line 23:
 
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}</math>.
 +
 
 +
==Solution 3==
 +
Notice that each team plays <math>N</math> games against each of the three other teams in its division. So that's <math>3N</math>.
 +
 
 +
Since each team plays <math>M</math> games against each of the four other teams in the other division, that's <math>4M</math>.
 +
 
 +
So <math>3N+4M=76</math>, with <math>M>4, N>2M</math>.
 +
 
 +
Let's start by solving this Diophantine equation. In other words, <math>M=\frac{76-4M}{3}</math>.
 +
 
 +
So <math>76-4M\equiv0 \pmod{3}</math>. Therefore, after reducing <math>76</math> to <math>1</math> and <math>-4M</math> to <math>2M</math> (we are doing things in <math>\pmod{3}</math>), we find that <math>M\equiv1 \pmod{3}</math>.
 +
 
 +
Since <math>M>4</math>, so the minimum possible value of <math>M</math> is <math>7</math>. However, remember that <math>N>2M</math>! To find the greatest possible value of M, we assume that <math>N=2M</math> and that is the upper limit of <math>M</math> (excluding that value because <math>N>2M</math>). Plugging <math>N=2M</math> in, <math>10M=76</math>. So <math>M<7.6</math>. Since you can't have <math>7.6</math> games, we know that we can only check <math>M=7</math> since we know that since <math>M>4, M<7.6, m\equiv1 (\pmod{3})</math>. After checking <math>M=7</math>, we find that it works.
 +
 
 +
So <math>M=7, N=16</math>. So each team plays 16 games against each team in its division. Select <math>\boxed{C}</math>.
 +
 
 +
This might be too complicated. But you should know what's happening by reading the [i]The Art of Problem Solving: Introduction to Number Theory[\i] 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.
 +
 
 +
~hastapasta
  
 
===Video Solutions===
 
===Video Solutions===

Revision as of 11:17, 30 March 2022

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, $M=6$ 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 $M=5,6,7$. 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 $\boxed{\textbf{(B)}~48}$.

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, $M=\frac{76-4M}{3}$.

So $76-4M\equiv0 \pmod{3}$. 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. Select $\boxed{C}$.

This might be too complicated. But you should know what's happening by reading the [i]The Art of Problem Solving: Introduction to Number Theory[\i] 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.

~hastapasta

Video Solutions

https://youtu.be/LiAupwDF0EY - Happytwin

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

https://youtu.be/HISL2-N5NVg?t=4968 - pi_is_3.14

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_24&oldid=173079"