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

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==Problem 20==
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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>  
 
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>  

Revision as of 16:12, 4 February 2016

Problem

A binary operation $\diamondsuit$ has the properties that $a\ \diamondsuit\ (b\ \diamondsuit\ c) = (a\ \diamondsuit\ b)\cdot c$ and that $a\ \diamondsuit\ a = 1$ for all nonzero real numbers $a, b$ and $c.$ (Here the dot $\ \cdot$ represents the usual multiplication operation.) The solution to the equation $2016\ \diamondsuit\ (6\ \diamondsuit\ x) = 100$ can be written as $\frac{p}{q},$ where $p$ and $q$ are relativelt prime positive integers. What is $p + q?$

$\textbf{(A)}\ 109\qquad\textbf{(B)}\ 201\qquad\textbf{(C)}\ 301\qquad\textbf{(D)}\ 3049\qquad\textbf{(E)}\ 33,601$

Solution

We can manipulate the given identities to arrive at a conclusion about the binary operator $\diamondsuit$. Substituting $b = c$ into the second identity yields $( a\ \diamondsuit\ b) \cdot b = a\ \diamondsuit\ (b\ \diamondsuit\  b) = a\ \diamondsuit\  1 = a\ \diamondsuit\ ( a\ \diamondsuit\ a) = ( a\ \diamondsuit\ a) \cdot a = a$. Hence, $( a\ \diamondsuit\ b) \cdot b = a,$ or, dividing both sides of the equation by $a,$ $( a\ \diamondsuit\ b) = \frac{a}{b}.$

Hence, the given equation becomes $\frac{2016}{\frac{60}{x}} = 100$. Solving yields $x=\frac{100}{336} = \frac{25}{84},$ so the answer is $25 + 84 = \boxed{\textbf{(A) }109.}$

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