2019 AMC 10B Problem 24

The target $4 + 2^{-20}$ suggests measuring the distance from $4$. Let $y_n = x_n - 4$, so $y_0 = 1$. Then
$$y_{n+1} = \frac{x_n^2 + 5x_n + 4}{x_n + 6} - 4 = \frac{x_n^2 + x_n - 20}{x_n + 6} = \frac{(x_n - 4)(x_n + 5)}{x_n + 6} = y_n \cdot \frac{y_n + 9}{y_n + 10}.$$

For $0 < y_n \le 1$ the factor $\frac{y_n + 9}{y_n + 10}$ lies between $\frac{9}{10}$ (at $y_n = 0$) and $\frac{10}{11}$ (at $y_n = 1$), so $y_n$ stays positive, decreases, and satisfies
$$\left(\frac{9}{10}\right)^n \le y_n \le \left(\frac{10}{11}\right)^n.$$

We need the first $m$ with $y_m \le 2^{-20}$.

• Lower bound on $m$: $\left(\frac{10}{9}\right)^6 = \frac{1000000}{531441} < 2$, so $\left(\frac{9}{10}\right)^{120} > 2^{-20}$, hence $y_{120} > 2^{-20}$ and $m > 120$.
• Upper bound on $m$: $1.1^8 \approx 2.14 > 2$, so $\left(\frac{10}{11}\right)^{160} < 2^{-20}$, hence $y_{160} \le 2^{-20}$ and $m \le 160$.

Thus $120 < m \le 160$, which lies inside $[81, 242]$.

The answer is $\boxed{\textbf{(C)}\ [81,242]}$.

A rational recursion that converges to a fixed point is best studied through the error term $x_n - L$; factoring reveals the per-step contraction ratio. Since the ratio is trapped in a narrow band, the error decays geometrically at a known rate, and two crude power comparisons ($1.1^8 > 2$ and $(10/9)^6 < 2$) squeeze $m$ into a range far narrower than the answer intervals. The intervals are powers of $3$ apart precisely so that rough bounds suffice.

• Trying to compute terms exactly, or bounding with only one side of the ratio and landing in the wrong interval.
• Working with $x_n$ itself and thinking it approaches $4$ "quickly" (guessing (A)) because $x_1 = \frac{54}{11} \approx 4.9$ already looks close.