2016 AMC 12A Problems/Problem 20

Revision as of 23:52, 4 February 2016 by Brone (talk | contribs) (Solution: typo)

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 $b,$ $( a\ \diamondsuit\ b) = \frac{a}{b}.$

Hence, the given equation becomes $\frac{2016}{\frac{6}{x}} = 100$. Solving yields $x=\frac{100}{336} = \frac{25}{84},$ so the answer is $25 + 84 = \boxed{\textbf{(A) }109.}$

See Also

2016 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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