2006 AMC 10B Problem 18

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 $\boxed{\textbf{(E)}\ 3}$.

A two-term recurrence is periodic as soon as a consecutive pair of terms reappears, because everything after is forced. For an index like $2006$ the intent is clearly "find the cycle," so compute terms until the starting pair returns, then reduce the index mod the period. Structurally, $a_n = a_{n-1}/a_{n-2}$ gives $a_n = 1/a_{n-3}$, hence $a_n = a_{n-6}$.

• Stopping at a_3 or a_4 and guessing, or computing 2006 mod 6 incorrectly (it is 2, not 0 or 4).
• Taking remainder $0$ to mean $a_0$ rather than $a_6$ when the index happens to be a multiple of the period.