# Difference between revisions of "2016 AMC 10A Problems/Problem 23"

## 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 $\cdot$ represents multiplication). The solution to the equation $2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100$ can be written as $\tfrac{p}{q}$, where $p$ and $q$ are relatively 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 see that $a \diamond a = 1$, and think of division. Testing, we see that the first condition $a \diamond (b \diamond c) = (a \diamond b) \cdot c$ is satisfied, because $\frac{a}{\frac{b}{c}} = \frac{a}{b} \cdot c$. Therefore, division is the operation $\diamond$. Solving the equation, $\frac{2016}{\frac{6}{x}} = \frac{2016}{6} \cdot x = 336x = 100$, so $x=\frac{100}{336} = \frac{25}{84}$, so the answer is $25 + 84 = \boxed{(A)109}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 