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"

(→‎Solution 2)
 
(58 intermediate revisions by 31 users not shown)
Line 1: Line 1:
−
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?
+
== Problem ==
  
−
<math>
+
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?
−
\textbf{(A) } 36 \qquad
 
−
\textbf{(B) } 48 \qquad
 
−
\textbf{(C) } 54 \qquad
 
−
\textbf{(D) } 60 \qquad
 
−
\textbf{(E) } 72
 
−
</math>
 
  
−
===Solution 1===
+
<math>\textbf{(A) } 36 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 54 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 72 </math>
−
Note that the equation rewrites to <math>3N+4M=76</math>.
 
  
−
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.
+
==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>.
  
−
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>.
+
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>.
  
−
===Solution 2===
+
Next, <math>M=6</math> does not work because <math>52</math> is not divisible by <math>3</math>.
−
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>  
 
  
−
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>.
+
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.
  
−
Next <math>M=6</math>, does not work because <math>52</math> is not divisible by <math>3</math>
+
<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.
  
−
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.
+
==Solution 2==
−
 
 
−
===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>.
−
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>.
 
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>N=\frac{76-4M}{3}</math>.
 +
 +
So <math>76-4M\equiv0 \pmod{3}</math> (remember: <math>M</math> must be divisible by 3 for <math>N</math> to be an integer!). 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 will work.
 +
 +
So <math>M=7, N=16</math>. So each team plays 16 games against each team in its division. Since they are asking for games in it division, which equals <math>3n = 48</math>. Select <math>\boxed{\textbf{(B)}~48}</math>.
 +
 +
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 and little kids 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==
 
==See Also==

Latest revision as of 23:35, 23 December 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 $M=5,6,7$. 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, $N=\frac{76-4M}{3}$.

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 will work.

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 and little kids 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
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=237700"