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. Observe that $x_{n+1} - 4 = \frac{x_n^2 + 5x_n + 4 - 4(x_n+6)}{x_n+6} = \frac{(x_n - 4)(x_n+5)}{x_n+6}$ so (since $x_n$ is clearly positive for all $n$ , from the initial definition), $x_{n+1} > 4$ if and only if $x_{n} > 4$ .

We similarly prove that $x_n$ is decreasing, since $x_{n+1} - x_n = \frac{x_n^2 + 5x_n + 4 - x_n(x_n+6)}{x_n+6} = \frac{4-x_n}{x_n+6} < 0$

Now we need to estimate the value of $x_{n+1}-4$ , which we can do using the rearranged equation $x_{n+1} - 4 = (x_n-4)\cdot\frac{x_n + 5}{x_n+6}$ Since $x_n$ is decreasing, $\frac{x_n + 5}{x_n+6}$ is clearly also decreasing, so we have $\frac{9}{10} < \frac{x_n + 5}{x_n+6} \le \frac{10}{11}$ and $\frac{9}{10}(x_n-4) < x_{n+1} - 4 \le \frac{10}{11}(x_n-4)$

This becomes $\left(\frac{9}{10}\right)^n = \left(\frac{9}{10}\right)^n \left(x_0-4\right) < x_{n} - 4 \le \left(\frac{10}{11}\right)^n \left(x_0-4\right) = \left(\frac{10}{11}\right)^n$ The problem thus reduces to finding the least value of $n$ such that $\left(\frac{9}{10}\right)^n < x_{n} - 4 \le \frac{1}{2^{20}} \text{ and } \left(\frac{10}{11}\right)^{n-1} > x_{n-1} - 4 > \frac{1}{2^{20}}$

Taking logarithms, we get $n \ln \frac{9}{10} < -20 \ln 2$ and $(n-1)\ln \frac{10}{11} > -20 \ln 2$ , i.e.

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

As approximations, we can use $\ln\frac{10}{9} \approx \frac{1}{9}$ , $\ln\frac{11}{10} \approx \frac{1}{10}$ , and $\ln 2\approx 0.7$ . These allow us to estimate that $126 < n < 141$ which gives the answer as $\boxed{\textbf{(C) } [81,242]}$ .

Solution by mathsuper(丹神)

Solution 2

The condition where $x_m\leq 4+\frac{1}{2^{20}}$ gives the motivation to make a substitution to change the equilibrium from $4$ to $0$ . We can substitute $x_n = y_n + 4$ to achieve that.

Now, we need to find the smallest value of $m$ such that $y_m\leq \frac{1}{2^{20}}$ given that $y_0 = 1$ and the recursion $y_{n+1} = \frac{y_n^2 + 9y_n}{y_n + 10}$ .

Using wishful thinking, we can simplify the recursion as follows:

$y_{n+1} = \frac{y_n^2 + 9y_n + y_n - y_n}{y_n + 10}$

$y_{n+1} = \frac{y_n(y_n + 10) - y_n}{y_n + 10}$

$y_{n+1} = y_n - \frac{y_n}{y_n + 10}$

$y_{n+1} = y_n(1 - \frac{1}{y_n + 10})$

The recursion looks like a geometric sequence with the ratio changing slightly after each term. Notice from the recursion that the $y_n$ sequence is strictly decreasing, so all the terms after $y_0$ will be less than 1. Also, notice that all the terms in sequence will be positive. Both of these can be proven by induction.

With both of those observations in mind, $\frac{9}{10} < 1 - \frac{1}{y_n + 10} \leq \frac{10}{11}$ . Combining this with the fact that the recursion resembles a geometric sequence, we conclude that $(\frac{9}{10})^2 < y_n \leq (\frac{10}{11})^2$ .

Note: Solution is incomplete

2019 AMC 10B (Problems • Answer Key • Resources)
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2019 AMC 10B Problems/Problem 24

Contents

Problem

Solution 1

Solution 2

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