2019 AMC 12B Problem 22

Let $a_n=x_n-4.$ A short computation gives $$\begin{gathered} a_{n+1}=x_{n+1}-4 \\ =\dfrac{(x_n+5)(x_n-4)}{x_n+6} \\ =a_n\cdot\dfrac{x_n+5}{x_n+6}. \end{gathered}$$ Starting from $a_0=1,$ the terms stay positive and decrease. Because $x_n$ decreases from $5$ toward $4,$ each ratio $\dfrac{x_n+5}{x_n+6}$ lies strictly between $\dfrac{9}{10}$ and $\dfrac{10}{11}.$ Hence $a_m$ is squeezed between $\left(\dfrac{9}{10}\right)^m$ and $\left(\dfrac{10}{11}\right)^m.$ Solving $a_m\le2^{-20}$ puts $m$ between about $132$ and $146,$ which lies in $[81,242].$ Thus, C is the correct answer.