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

Revision as of 11:50, 29 May 2019

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 $\binom{3}{2}N$ games in their division and $4(M)$ 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 $M=7$ this does work 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 9: / Line 9: @@
 </math>
-===Solution 1===
+====Solution 1====
 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>

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

Revision as of 11:50, 29 May 2019

Solution 1

Solution 2

See Also