2019 AMC 10B Problem 24

Let $a_n=x_n-4$. Then $a_0=1$, and simplifying the recurrence gives $$a_{n+1}=\frac{a_n(a_n+9)}{a_n+10}.$$ As long as $0<a_n\le1$, this implies $$\frac9{10}a_n\le a_{n+1}\le\frac{10}{11}a_n.$$ By induction, $\left(\dfrac9{10}\right)^n\le a_n\le\left(\dfrac{10}{11}\right)^n$. For $n=80$, $\left(\dfrac{10}{9}\right)^{80}<2^{20}$, so $\left(\dfrac9{10}\right)^{80}>2^{-20}$, and therefore $m>80$. Also $\left(\dfrac{11}{10}\right)^8>2$, so $\left(\dfrac{10}{11}\right)^{160}<2^{-20}$, which gives $m\le160$. Hence $81\le m\le160$, so $m$ lies in $[81,242]$. Thus, C is the correct answer.