# 2012 AIME II Problems/Problem 11

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## Problem 11

Let $f_1(x) = \frac23 - \frac3{3x+1}$, and for $n \ge 2$, define $f_n(x) = f_1(f_{n-1}(x))$. The value of $x$ that satisfies $f_{1001}(x) = x-3$ can be expressed in the form $\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

## Solution

After evaluating the first few values of $f_k (x)$, we obtain $f_4(x) = f_1(x) = \frac{2}{3} - \frac{3}{3x+1} = \frac{6x-7}{9x+3}$. Since $1001 \equiv 2 \mod 3$, $f_{1001}(x) = f_2(x) = \frac{3x+7}{6-9x}$. We set this equal to $x-3$, i.e.

$\frac{3x+7}{6-9x} = x-3 \Rightarrow x = \frac{5}{3}$. The answer is thus $5+3 = \boxed{008}$.