2016 AMC 10A Problems/Problem 23

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

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 $25 + 84 = \boxed{109}$ (A)

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2016 AMC 10A Problems/Problem 23

Solution