# 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 $\boxed{C}$

## Solution 2

Making the reasonable assumption that $x_n$ approaches $4$, we can translate $x$ down by $4$ to obtain a more simple sequence $a_n=x_n-4$ that should approach $0$.

Substitution of $(a_{n}+4)$ for $(x_{n})$ and $(a_{n+1}+4)$ for $(x_{n+1})$ in the definition of $x_{n+1}$ leads to $a_{n+1}+4 = \frac{(a_{n} + 8)(a_{n}+5)}{a_{n}+10} = \frac{a_{n}^2 + 13a_{n}+40}{a_{n}+10} \implies a_{n+1} = a_{n}(\frac{a_{n}+9}{a_{n}+10}) \implies \frac{a_{n+1}}{a_{n}} = \frac{a_{n}+9}{a_{n}+10}$

The ratio of consecutive terms is thus always positive and less than 1 (because $a_0$ is positive). This means that the largest possible value for $a_n$ is 1 and that no value of $a_n$ can be less than or equal to 0.

Plugging the extrema of $a_n$ back into the ratio shows that $\frac{9}{10}<\frac{a_{n+1}}{a_{n}}\leq\frac{10}{11}$ for all $n$.

For $(n>0)$, we can bound $a_{n}$ by applying this rule recursively  : $(\frac{9}{10})^n < a_{n} \leq (\frac{10}{11})^n$

Therefore, $a_{n}$ is always less than $(\frac{1}{2^{20}})$ when $(\frac{10}{11})^n<\frac{1}{2^{20}}\implies n>20log_{\frac{11}{10}}{2}$

and $a_{n}$ is never less than $(\frac{1}{2^{20}})$ when $(\frac{9}{10})^n>\frac{1}{2^{20}}\implies n<20log_{\frac{10}{9}}{2}$

The first integer $m$ such that $a_{m}\leq\frac{1}{2^{20}}$ must therefore lie in the interval $\left[\left\lfloor 20log_{\frac{10}{9}}{2}\right\rfloor+1, \left\lceil 20log_{\frac{11}{10}}{2}\right\rceil\right]$

Both of these can be quickly estimated at c. $140$, so the answer must be $\boxed{C}$. (actual values are $132$ and $147$)

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 