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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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| − | ==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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