2015 AMC 8 Problems/Problem 24: Difference between revisions

Revision as of 10:39, 8 January 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 $\boxed{\textbf{(B)}~48}$

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

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

@@ Line 8: / Line 8: @@
 ===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>.
 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>.
-Next, <math>M=6</math> does not work because <math>52</math> is not divisible by <math>3</math>
+Next, <math>M=6</math> does not work because <math>52</math> is not divisible by <math>3</math>.
 We try <math>M=7</math> <math>does</math> work by giving <math>N=16,~M=7</math> and thus <math>3\times 16=\boxed{\textbf{(B)}~48}</math> games in their division.

Page

Toolbox

Search