2006 AMC 10B Problem 18

1. The index $2006$ is too big to reach step by step, so look for a repeating pattern. Each term is the one before it divided by the one before that.
2. Work out terms:
$$a_1 = 2,\quad a_2 = 3,\quad a_3 = \frac{3}{2},\quad a_4 = \frac{3/2}{3} = \frac{1}{2},\quad a_5 = \frac{1/2}{3/2} = \frac{1}{3},$$
$$a_6 = \frac{1/3}{1/2} = \frac{2}{3},\quad a_7 = \frac{2/3}{1/3} = 2,\quad a_8 = \frac{2}{2/3} = 3.$$
3. The pair $2, 3$ has come back at $a_7, a_8$. Each term depends only on the two before it, so from here the sequence repeats exactly. The pattern is $6$ terms long:
$$2,\ 3,\ \tfrac32,\ \tfrac12,\ \tfrac13,\ \tfrac23,\ 2,\ 3,\ \ldots$$
4. Divide: $2006 = 6 \cdot 334 + 2$. After $334$ full rounds of $6$, we are at position $2$ of the pattern.
5. So $a_{2006} = a_2 = 3$.

The answer is $\boxed{\textbf{(E)}\ 3}$.

When each term is built from the two before it, the sequence must repeat once two neighboring terms come back. A huge index like $2006$ is a strong hint to look for that cycle. Find the length of the cycle, then use the remainder to find the position inside it.

See why the cycle has length $6$. Using the rule twice, $a_n = \dfrac{a_{n-1}}{a_{n-2}} = \dfrac{a_{n-2}/a_{n-3}}{a_{n-2}} = \dfrac{1}{a_{n-3}}$. So each term is $1$ over the term three places back, and flipping twice gets back to the start: $a_n = a_{n-6}$.

Generate terms until a repeat appears:
$$a_1 = 2,\quad a_2 = 3,\quad a_3 = \frac{3}{2},\quad a_4 = \frac{3/2}{3} = \frac{1}{2},\quad a_5 = \frac{1/2}{3/2} = \frac{1}{3},\quad a_6 = \frac{1/3}{1/2} = \frac{2}{3},$$
$$a_7 = \frac{2/3}{1/3} = 2,\quad a_8 = \frac{2}{2/3} = 3.$$
Since $a_7 = a_1$ and $a_8 = a_2$, and each term is determined by the two before it, the sequence repeats with period $6$: $a_{n+6} = a_n$ for all $n$.

Now $2006 = 6 \cdot 334 + 2$, so $a_{2006} = a_2 = 3$.

The answer is $\boxed{\textbf{(E)}\ 3}$.