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: $x_{n+1} - x_n = \frac{x_n^2 + 5x_n + 4 - x_n(x_n+6)}{x_n+6} = \frac{4-x_n}{x_n+6} < 0.$

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 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 approximations allow us to estimate $126 < n < 141,$ which gives $\boxed{\textbf{(C) } [81,242]}$ .

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 $y_0 = 1$ .

Factoring the recursion $x_{n+1} = \frac{x_n^2 + 5x_n+4}{x_n + 6}$ , we get:

$x_{n+1}=\dfrac{(x_n + 4)(x_n + 1)}{x_n + 6} \Rightarrow y_{n+1}+4=\dfrac{(y_n+8)(y_n+5)}{y_n+10}$

$y_{n+1}+4=\dfrac{y_n^2+13y_n+40}{y_n+10} = \dfrac{y_n^2+9y_n +(4y_n+40)}{y_n+10}$

$y_{n+1}+4=\dfrac{y_n^2+9y_n}{y_n+10} + 4$

$y_{n+1}=\dfrac{y_n^2+9y_n}{y_n+10}$ .

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

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

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

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

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

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 $\left(\frac{9}{10}\right)^n < y_n \leq \left(\frac{10}{11}\right)^n.$

$\frac{9}{10}$ is approximately equal to $\frac{10}{11}$ and the ranges that the answer choices give us are generous, so we should use either $\frac{9}{10}$ or $\frac{10}{11}$ to find a rough estimate for $m$ .

Since $\dfrac{1}{2}=0.5$ , that means $\frac{1}{\sqrt{2}}=2^{-\frac{1}{2}} \approx 0.7$ . Additionally, $\left(\frac{9}{10}\right)^3=0.729$

Therefore, we can estimate that $2^{-\frac{1}{2}} < y_3$ .

Raising both sides to the 40th power, we get $2^{-20} < (y_3)^{40}$

But $y_3 = (y_0)^3$ , so $2^{-20} < (y_0)^{120}$ and therefore, $2^{-20} < y_{120}$ .

This tells us that $m$ is somewhere around 120, so our answer is $\boxed{\textbf{(C) } [81,242]}$ .

Solution 3

Since the choices are rather wide ranges, we can use approximation to make it easier. Notice that $x_{n+1} - x_n = \frac{4-x_n}{x_n+6}$ And $x_0 =5$ , we know that $x_n$ is a declining sequence, and as it get close to 4 its decline will slow, never falling below 4. So we'll use 4 to approximate $x_n$ in the denominator so that we have a solvable difference equation: $x_{n+1} - x_n = \frac{4-x_n}{10}$ $x_{n+1} = \frac{9}{10}x_n + \frac{2}{5}$ Solve it with $x_0 = 5$ , we have $x_n = 4 + (\frac{9}{10})^n$ Now we wish to find $n$ so that $(\frac{9}{10})^n \approx \frac{1}{2^{20}}$ $n \approx \frac{\log{2^{20}}}{\log{10}-\log9} \approx \frac{20*0.3}{0.05} = 120$ Since 120 is safely within the range of [81,242], we have the answer. $\boxed{\textbf{(C) } [81,242]}$ .

-Mathdummy

Solution 4 (no logs)

We start by simplifying the recursion: $x_{n+1} = x_n + \frac{4-x_n}{x_n+6}$ . While $x_n > 4$ , $x_n$ is a decreasing sequence. Let $x_n = 4 + f_n$ . Then $x_{n} - x_{n+1} = \frac{x_n-4}{x_n+6} = \frac{f_n}{10 + f_n} \approx \frac{f_n}{10}.$ Now notice that we want to find $m$ , such that $x_m \leq 4 + \frac{1}{2^{20}}$ , and $x_0 = 4 + \frac{1}{2^0}$ . We are going to find an approximate number of steps to go from $4 + \frac{1}{2^i}$ to $4 + \frac{1}{2^{i+1}}$ . If $x_j = 4 + \frac{1}{2^i}$ , then $f_j = \frac{1}{2^i}$ and if $x_k = 4 + \frac{1}{2^{i+1}}$ , then $f_k = \frac{1}{2^{i+1}}$ . Then $x_j - x_k = \frac{1}{2^{i+1}} \approx \frac{f_j}{10} + \frac{f_{j+1}}{10} + ... + \frac{f_{k-1}}{10}.$ Furthermore, $\frac{1}{2^i} = f_j > f_{j+1} > f_{j+2} > ... > f_{k-1} > f_k = \frac{1}{2^{i+1}}.$ Therefore, $(k-j) \cdot \frac{\frac{1}{2^i}}{10} > \frac{f_j}{10} + \frac{f_{j+1}}{10} + ... + \frac{f_{k-1}}{10} > (k-j) \cdot \frac{\frac{1}{2^{i+1}}}{10},$ so $5 < k-j < 10$ . Therefore, to go from $f_0 = \frac{1}{2^0}$ to $f_m = \frac{1}{2^{20}}$ , we will need to do this 20 times, which means $100 < m - 0 < 200$ , which is within the range of [81,242], so we have the answer. $\boxed{\textbf{(C) } [81,242]}$ .

-alligator112

Solution 5

As in Solution 1, we show that all $x_n>4$ by induction.

For our base case, we simply know that $x_0>4$ . Then, for our inductive step, assume that $x_k>4$ for some $k$ . Then we may let $x_k=4+a$ , where $a>0$ . Plugging this into the equation yields

$x_{k+1}=\frac{(4+a)^2+5(4+a)+4}{(4+a)+6}$

$=\frac{a^2+13a+40}{a+10}$

$=a+3+\frac{10}{a+10}$

[INCOMPLETE]

Remark

The actual value of $m$ is $m=133$ .

Video Solution

https://www.youtube.com/watch?v=toHqTtcCcJQ

~MathProblemSolvingSkills.com

Video Solution

Video Solution: https://www.youtube.com/watch?v=Ok7bYOdiF6M

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

Contents

Problem

Solution 1

Solution 2

Solution 3

Solution 4 (no logs)

Solution 5

Remark

Video Solution

Video Solution

See Also