# 2019 AMC 10B Problems/Problem 24

The following problem is from both the 2019 AMC 10B #24 and 2019 AMC 12B #22, so both problems redirect to this page.

## Problem

Define a sequence recursively by $x_0=5$ and $$x_{n+1}=\frac{x_n^2+5x_n+4}{x_n+6}$$ for all nonnegative integers $n.$ Let $m$ be the least positive integer such that $$x_m\leq 4+\frac{1}{2^{20}}.$$In which of the following intervals does $m$ lie?

$\textbf{(A) } [9,26] \qquad\textbf{(B) } [27,80] \qquad\textbf{(C) } [81,242]\qquad\textbf{(D) } [243,728] \qquad\textbf{(E) } [729,\infty)$

## Solution 1

We first prove that $x_n > 4$ for all $n \ge 0$ by induction from $$x_{n+1} - 4 = \frac{x_n^2 + 5x_n + 4 - 4(x_n+6)}{x_n+6} = \frac{(x_n - 4)(x_n+5)}{x_n+6}$$ and then prove $x_n$'s are decreasing by $$x_{n+1} - x_n = \frac{x_n^2 + 5x_n + 4 - x_n(x_n+6)}{x_n+6} = \frac{4-x_n}{x_n+6} < 0$$ Now we need to estimate the value of $x_{n+1}-4$ by $$x_{n+1} - 4 = (x_n-4)\cdot\frac{x_n + 5}{x_n+6}$$ since $x_n$'s are decreasing, $\frac{x_n + 5}{x_n+6}$ are also decreasing, so we have $$\frac{9}{10} < \frac{x_n + 5}{x_n+6} \le \frac{10}{11}$$ and $$\frac{9}{10}(x_n-4) < x_{n+1} - 4 \le \frac{10}{11}(x_n-4)$$ which leads to $$(\frac{9}{10})^n = (\frac{9}{10})^n (x_0-4) < x_{n} - 4 \le (\frac{10}{11})^n (x_0-4) = (\frac{10}{11})^n$$ The problem requires us to find the value of $n$ such that $$(\frac{9}{10})^n < x_{n} - 4 \le \frac{1}{2^{20}} \text{ and } (\frac{10}{11})^{n-1} > x_{n-1} - 4 > \frac{1}{2^{20}}$$ using natural logarithm, we need $n \ln \frac{9}{10} < -20 \ln 2$ and $(n-1)\ln \frac{10}{11} > -20 \ln 2$, or

$$n > \frac{20\ln 2}{\ln\frac{10}{9}} \text{ and } n-1 < \frac{20\ln 2}{\ln\frac{11}{10}}$$

As estimations, $\ln\frac{10}{9} \approx 1/9$ and $\ln\frac{11}{10} \approx 1/10$, $\ln 2\approx 0.7$ we can estimate that $$126 < n < 141$$ Choose $\boxed{C}$