Difference between revisions of "2013 AMC 12A Problems/Problem 2"

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Therefore we can deduce that when they scored an odd number of runs, they lost by one, and when they scored an even number of runs, they won by twice as much.
 
Therefore we can deduce that when they scored an odd number of runs, they lost by one, and when they scored an even number of runs, they won by twice as much.
  
Therefore, the total runs by the opponent is <math>(2+4+6+8+10)+(1+2+3+4+5) = 45</math>
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Therefore, the total runs by the opponent is <math>(2+4+6+8+10)+(1+2+3+4+5) = 45</math>, which is <math>C</math>

Revision as of 00:52, 8 February 2013

To score twice as many runs as their opponent, the softball team must have scored an even number.

Therefore we can deduce that when they scored an odd number of runs, they lost by one, and when they scored an even number of runs, they won by twice as much.

Therefore, the total runs by the opponent is $(2+4+6+8+10)+(1+2+3+4+5) = 45$, which is $C$

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