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| − | == Problem ==
| + | Big balck men |
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| − | 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>\textbf{(A) } 36 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 54 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 72 </math>
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| − | ==Solution 1==
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| − | 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>.
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| − | Since <math>M>4</math> we start by trying M=5. 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>.
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| − | We try <math>M=7</math> does work by giving <math>N=16</math> ,<math>~M=7</math> and thus <math>3\times 16=\boxed{\textbf{(B)}~48}</math> games in their division.
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| − | <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.
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| − | ==Solution 2==
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| − | <math>76=3N+4M > 10M</math>, giving <math>M \le 7</math>.
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| − | Since <math>M>4</math>, we have <math>M=5,6,7</math>.
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| − | 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>.
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| − | This gives <math>3N=48</math>, as desired. The answer is <math>\boxed{\textbf{(B)}~48}</math>.
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| − | ==Solution 3==
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| − | Notice that each team plays <math>N</math> games against each of the three other teams in its division. So that's <math>3N</math>.
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| − | Since each team plays <math>M</math> games against each of the four other teams in the other division, that's <math>4M</math>.
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| − | So <math>3N+4M=76</math>, with <math>M>4, N>2M</math>.
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| − | Let's start by solving this Diophantine equation. In other words, <math>N=\frac{76-4M}{3}</math>.
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| − | So <math>76-4M\equiv0 \pmod{3}</math> (remember: <math>M</math> must be divisible by 3 for <math>N</math> to be an integer!). Therefore, after reducing <math>76</math> to <math>1</math> and <math>-4M</math> to <math>2M</math> (we are doing things in <math>\pmod{3}</math>), we find that <math>M\equiv1 \pmod{3}</math>.
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| − | Since <math>M>4</math>, so the minimum possible value of <math>M</math> is <math>7</math>. However, remember that <math>N>2M</math>! To find the greatest possible value of M, we assume that <math>N=2M</math> and that is the upper limit of <math>M</math> (excluding that value because <math>N>2M</math>). Plugging <math>N=2M</math> in, <math>10M=76</math>. So <math>M<7.6</math>. Since you can't have <math>7.6</math> games, we know that we can only check <math>M=7</math> since we know that since <math>M>4, M<7.6, M\equiv1 \pmod{3}</math>. After checking <math>M=7</math>, we find that it will work.
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| − | So <math>M=7, N=16</math>. So each team plays 16 games against each team in its division. Since they are asking for games in it division, which equals <math>3n = 48</math>. Select <math>\boxed{\textbf{(B)}~48}</math>.
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| − | This might be too complicated. But you should know what's happening if you read the ''The Art of Problem Solving: Introduction to Number Theory'' by Mathew Crawford. Notice how I used chapter 12's ideas of basic modular arithmetic operations and chapter 14's ideas of solving linear congruences. Remember: the Introduction Series books by AoPS are for 6th-10th graders! So make sure to read the curriculum books!
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| − | This goes same for high school students as well, and especially for those who want to continue on their path of AIME/USA(J)MO.
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| − | ~hastapasta
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| − | ===Video Solution by OmegaLearn===
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| − | https://youtu.be/HISL2-N5NVg?t=4968
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| − | ===Video Solutions===
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| − | https://youtu.be/LiAupwDF0EY - Happytwin
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| − | https://www.youtube.com/watch?v=bJSWtw91SLs - Oliver Jiang
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| − | https://youtu.be/0ZMDsuOYGqE
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| − | ~savannahsolver
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| − | ==See Also==
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| − | {{AMC8 box|year=2015|num-b=23|num-a=25}}
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| − | {{MAA Notice}}
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