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

Revision as of 17:43, 8 January 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, $M=6$ 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 $\boxed{\textbf{(B)}~48}$ .

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

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

@@ Line 24: / Line 24: @@
 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. Select <math>\boxed{B}</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
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