2016 AMC 10A Problem 23

The strange operation is probably a familiar one in disguise. Use the rules to find out which.

1. Make the second rule appear inside the first. The first rule has $b \diamondsuit c$ in it. If we choose $c = b$, that becomes $b \diamondsuit b$, which is $1$. So the first rule turns into
$$a \diamondsuit 1 = (a \diamondsuit b) \cdot b.$$
Dividing by $b$:
$$a \diamondsuit b = \frac{a \diamondsuit 1}{b}.$$
2. Find $a \diamondsuit 1$. Now choose $b = a$ in this last line. The left side is $a \diamondsuit a = 1$, so
$$1 = \frac{a \diamondsuit 1}{a}, \qquad\text{which gives}\qquad a \diamondsuit 1 = a.$$
3. So the operation is division:
$$a \diamondsuit b = \frac{a}{b}.$$
Check both rules. Second rule: $\tfrac{a}{a} = 1$. First rule: $a \diamondsuit (b \diamondsuit c) = \dfrac{a}{b/c} = \dfrac{ac}{b}$, and $(a \diamondsuit b) \cdot c = \dfrac{a}{b} \cdot c = \dfrac{ac}{b}$. Both hold.
4. Solve the equation. $6 \diamondsuit x = \tfrac{6}{x}$, so
$$2016 \diamondsuit \frac{6}{x} = 2016 \div \frac{6}{x} = \frac{2016x}{6} = 336x.$$
5. Set $336x = 100$: $x = \tfrac{100}{336} = \tfrac{25}{84}$ (divide top and bottom by $4$).
6. $25 = 5^2$ and $84 = 2^2 \cdot 3 \cdot 7$ share no factor, so $p = 25$, $q = 84$ and $p + q = 109$.

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

An operation given only by rules can be pinned down by plugging in clever values, especially ones that make a known rule show up (here $b \diamondsuit b = 1$). Once you think you know the operation, test it against every rule before using it. Guessing is fine too: "$a \diamondsuit a = 1$" already suggests division.

Use the first rule directly on the equation, without finding $6 \diamondsuit x$: it says $2016 \diamondsuit (6 \diamondsuit x) = (2016 \diamondsuit 6) \cdot x$. With division, $2016 \diamondsuit 6 = 336$, so again $336x = 100$.