Difference between revisions of "2015 AMC 8 Problems/Problem 24"
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We try <math>M=7</math> this does work giving <math>N=16,~M=7</math> and thus <math>3\times 16=\boxed{\textbf{(B)},~48}</math> games in their division. | We try <math>M=7</math> this does work giving <math>N=16,~M=7</math> and thus <math>3\times 16=\boxed{\textbf{(B)},~48}</math> games in their division. | ||
+ | |||
+ | ===Solution 3=== | ||
+ | <math>76=3N+4M > 10M</math>, giving <math>M \le 7</math>. | ||
+ | Since <math>M>4</math>, we have <math>M=5,6,7</math> | ||
+ | Since <math>4M</math> is <math>1</math> <math>\pmod{3}</math>, we must have <math>M</math> equal to <math>1</math> <math>\pmod{3}</math>, so <math>M=7</math>. | ||
+ | |||
+ | This gives <math>3N=48</math>, as desired. The answer is <math>\boxed{\textbf{(B)}~48}</math>. |
Revision as of 17:05, 25 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?
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
.