# 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 .