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

Revision as of 12:21, 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, $N=\frac{76-4M}{3}$ .

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. Select $\boxed{C}$ .

This might be too complicated. But you should know what's happening by reading 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.

~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

@@ Line 40: / Line 40: @@
 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.
+This might be too complicated. But you should know what's happening by reading 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.
 ~hastapasta

2015 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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