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

(Solution 2)
(Solution 1)
Line 8: Line 8:
\textbf{(E) } 72
\textbf{(E) } 72
===Solution 1===
Note that the equation rewrites to <math>3N+4M=76</math>.
Now remark that if <math>(m,n)</math> is a solution to this equation, then so is <math>(m+3,n-4)</math>.  This is because <cmath>3(n-4)+4(m+3)=3n-12+4m+12=3n+4m=76.</cmath> Thus, we can now take an "edge case" solution and work upward until both conditions (<math>N>2M</math> and <math>M>4</math>) are met.
We see by inspection that <math>(M,N)=(1,24)</math> is a solution.  By the above work, we can easily deduce that <math>(4,20)</math> and <math>(7,16)</math> are solutions.  The last one is the intended answer (the next solution fails <math>N>2M</math>) so our answer is <math>3N=\boxed{\textbf{(B)  }48}</math>.
===Solution 2===
===Solution 2===

Revision as of 21:47, 15 June 2016

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 2

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.

Solution 3

$76=3N+4M > 10M$, 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 $\boxed{\textbf{(B)}~48}$.

See Also

2015 AMC 8 (ProblemsAnswer KeyResources)
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

Invalid username
Login to AoPS