2015 AMC 8 Problems/Problem 24: Difference between revisions
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<math>76=3N+4M > 10M</math>, giving <math>M \le 7</math>. | <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>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>. | This gives <math>3N=48</math>, as desired. The answer is <math>\boxed{\textbf{(B)}~48}</math>. | ||
Revision as of 21:59, 1 December 2024
Problem
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 
game schedule. How many games does a team play within its own division?
Solution 1
On one team they play 
games in their division and 
games in the other. This gives 
.
Since we start by trying M=5. This doesn't work because 
is not divisible by 
.
Next, 
does not work because 
is not divisible by .
We try 
does work by giving 
,
and thus 
games in their division.
seems to work, until we realize this gives 
, but so this will not work.
Solution 2
, 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 | ||
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