Difference between revisions of "2015 AMC 8 Problems/Problem 24"
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− | + | == Problem == | |
− | <math> | + | A baseball league consists of two four-team divisions. Each team plays every other team in its division <math>N</math> games. Each team plays every team in the other division <math>M</math> games with <math>N>2M</math> and <math>M>4</math>. Each team plays a <math>76</math> game schedule. How many games does a team play within its own division? |
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− | </math> | ||
− | + | <math>\textbf{(A) } 36 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 54 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 72 </math> | |
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− | + | ==Solutions== | |
− | + | ===Solution 1=== | |
+ | On one team they play <math>3N</math> games in their division and <math>4M</math> games in the other. This gives <math>3N+4M=76</math>. | ||
− | + | Since <math>M>4</math> we start by trying <math>M=5</math>. This doesn't work because <math>56</math> is not divisible by <math>3</math>. | |
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− | + | Next, <math>M=6</math> does not work because <math>52</math> is not divisible by <math>3</math>. | |
− | + | We try <math>M=7</math> <math>does</math> work by giving <math>N=16,~M=7</math> and thus <math>3\times 16=\boxed{\textbf{(B)}~48}</math> games in their division. | |
− | + | <math>M=10</math> seems to work, until we realize this gives <math>N=12</math>, but <math>N>2M</math> so this will not work. | |
− | ===Solution | + | ===Solution 2=== |
<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>. | 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> |
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+ | ===Video Solutions=== | ||
+ | https://youtu.be/LiAupwDF0EY - Happytwin | ||
+ | |||
+ | https://www.youtube.com/watch?v=bJSWtw91SLs - Oliver Jiang | ||
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+ | https://youtu.be/HISL2-N5NVg?t=4968 - pi_is_3.14 | ||
==See Also== | ==See Also== |
Latest revision as of 03:19, 14 January 2022
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?
Solutions
Solution 1
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 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
Video Solutions
https://youtu.be/LiAupwDF0EY - Happytwin
https://www.youtube.com/watch?v=bJSWtw91SLs - Oliver Jiang
https://youtu.be/HISL2-N5NVg?t=4968 - pi_is_3.14
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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