Difference between revisions of "2015 AMC 8 Problems/Problem 24"
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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. | 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{ | + | 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 13:28, 26 November 2015
A baseball league consists of two four-team divisions. Each team plays every other team in its division games. Each team plays every team in the other division
games with
and
. Each team plays a 76 game schedule. How many games does a team play within its own division?
Contents
Solution 1
Note that the equation rewrites to .
Now remark that if is a solution to this equation, then so is
. This is because
Thus, we can now take an "edge case" solution and work upward until both conditions (
and
) are met.
We see by inspection that is a solution. By the above work, we can easily deduce that
and
are solutions. The last one is the intended answer (the next solution fails
) so our answer is
.
Solution 2
On one team they play games in their division and
games in the other. This gives
Since we start by trying
. This doesn't work because
is not divisible by
.
Next , does not work because
is not divisible by
We try this does work giving
and thus
games in their division.
Solution 3
, giving
.
Since
, we have
Since
is
, we must have
equal to
, so
.
This gives , as desired. The answer is
.
See Also
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.