Difference between revisions of "2016 AMC 12A Problems/Problem 20"

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==Problem==
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#REDIRECT [[2016_AMC_10A_Problems/Problem_23]]
 
 
A binary operation <math>\diamondsuit </math> has the properties that <math>a\ \diamondsuit\ (b\ \diamondsuit\ c) = (a\ \diamondsuit\ b)\cdot c</math> and that <math>a\ \diamondsuit\ a = 1</math> for all nonzero real numbers <math>a, b</math> and <math>c.</math> (Here the dot <math>\ \cdot</math>  represents the usual multiplication operation.) The solution to the equation <math>2016\ \diamondsuit\ (6\ \diamondsuit\ x) = 100</math> can be written as <math>\frac{p}{q},</math> where <math>p</math> and <math>q</math> are relativelt prime positive integers. What is <math>p + q?</math>
 
 
 
<math>\textbf{(A)}\ 109\qquad\textbf{(B)}\ 201\qquad\textbf{(C)}\ 301\qquad\textbf{(D)}\ 3049\qquad\textbf{(E)}\ 33,601</math>
 
 
 
==Solution==
 
 
 
We can manipulate the given identities to arrive at a conclusion about the binary operator <math>\diamondsuit</math>.  Substituting <math>b = c</math> into the second identity yields <math>( a\ \diamondsuit\ b) \cdot b = a\ \diamondsuit\ (b\ \diamondsuit\  b) = a\ \diamondsuit\  1 = a\ \diamondsuit\ ( a\ \diamondsuit\ a) = ( a\ \diamondsuit\ a) \cdot a = a</math>.  Hence, <math>( a\ \diamondsuit\ b) \cdot b = a,</math> or, dividing both sides of the equation by <math>b,</math> <math>( a\ \diamondsuit\ b) = \frac{a}{b}.</math>
 
 
 
Hence, the given equation becomes <math>\frac{2016}{\frac{6}{x}} = 100</math>.  Solving yields <math>x=\frac{100}{336} = \frac{25}{84},</math> so the answer is <math>25 + 84 = \boxed{\textbf{(A) }109.}</math>
 
 
 
==See Also==
 
{{AMC12 box|year=2016|ab=A|num-b=19|num-a=21}}
 
{{MAA Notice}}
 

Latest revision as of 11:33, 5 February 2016

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