2015 AMC 8 Problems/Problem 24: Difference between revisions

Revision as of 17:44, 22 December 2024

Big balck men

2015 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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All AJHSME/AMC 8 Problems and Solutions

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 Big balck men
-==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 M=5. This doesn't work because <math>56</math> is not divisible by <math>3</math>.
-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> 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.
-<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.
 ==See Also==