2016 AMC 10A Problem 23

First identify the operation. In $a \diamondsuit (b \diamondsuit c) = (a \diamondsuit b)\cdot c$, set $c = b$; since $b \diamondsuit b = 1$ this gives
$$a \diamondsuit 1 = (a \diamondsuit b)\cdot b, \qquad\text{so}\qquad a \diamondsuit b = \frac{a \diamondsuit 1}{b}.$$
Now take $b = a$: $1 = a \diamondsuit a = \dfrac{a \diamondsuit 1}{a}$, hence $a \diamondsuit 1 = a$. Therefore
$$a \diamondsuit b = \frac{a}{b}.$$
(Check: $\frac{a}{b/c} = \frac{a}{b}\cdot c$ and $\frac aa = 1$, so division really satisfies both rules.)

Now solve. Using the first rule directly, $2016 \diamondsuit (6 \diamondsuit x) = (2016 \diamondsuit 6)\cdot x = \dfrac{2016}{6}\, x = 336x$. Setting $336x = 100$ gives
$$x = \frac{100}{336} = \frac{25}{84}.$$
Since $\gcd(25, 84) = 1$, $p + q = 25 + 84 = 109$.

The answer is $\boxed{\textbf{(A)}\ 109}$.

An abstract operation defined by identities is usually a familiar operation in disguise; plugging in values that trigger the known identity ($c = b$ to create $b \diamondsuit b = 1$) peels the definition away layer by layer. Once you suspect division, verify it against every given rule before using it. Note also that the associativity-like rule alone solves the equation without ever computing $6 \diamondsuit x$.

Guess $a \diamondsuit b = a/b$ from the two rules (the second rule says $a \diamondsuit a = 1$, like $a/a$), confirm it satisfies the first rule, then compute $2016 / (6/x) = 336x = 100$.

• Applying the rule as if diamond were multiplication or subtraction, or computing 6 diamond x = 6x instead of 6/x.
• Simplifying $\frac{100}{336}$ incompletely (e.g. $\frac{50}{168}$) and reporting a wrong $p + q$.