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"

(→‎Solutions)
(→‎Solution 2)
 
(2 intermediate revisions by 2 users not shown)
Line 5: Line 5:
 
<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>
  
−
==Solutions==
+
==Solution 1==
−
 
 
−
===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>.
  
Line 18: Line 16:
 
<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>.
−
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>.
+
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>.
−
 
−
==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 works.
 
−
 
−
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 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 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

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=223436"