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

This solution uses approximation to estimate ranges.

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

Next, we let $a_n=x_n-4$ , so then all $a_n>0$ . Replacing this in the formula results in $a_{n+1}+4=\frac{(a_n+4)^2+5(a_n+4)+4}{(a_n+4)+6}=\frac{a_n^2+13a_n+40}{a_n+10}$ , so then

$a_{n+1}=\frac{a_n^2+9a_n}{a_n+10}$

The problem statement asks us to find some $m$ such that $a_m\le\frac{1}{2^{20}}$ .

Then we let $b_n=\frac{1}{a_n}$ . Note that $a_n>0$ , so $b_n$ is never undefined. Replacing this in the formula results in $\frac{1}{b_{n+1}}=\frac{\left(\frac{1}{b_n}\right)^2+9\cdot\frac{1}{b_n}}{\frac{1}{b_n}+10}=\frac{9b_n+1}{10b_n^2+b_n}$ , so $b_{n+1}=\frac{10b_n^2+b_n}{9b_n+1}=\frac{10}{9}b_n-\frac{1}{81}+\frac{1}{729b_n+81}$ . Now we must find the least $m$ such that $b_m\ge2^{20}$ . Notice that only the $\frac{10}{9}n$ portion of the equation plays a significant role in determining the value of the entire value at higher values of $b_n$ , so we may simply let $b_{n+1}\approx\frac{10}{9}b_n$ .

Since $b_0=\frac{1}{a_n}=\frac{1}{x_0-4}=\frac{1}{5-4}=1$ , we know that $b_n\approx\left(\frac{10}{9}\right)^n=\left(\frac{100}{81}\right)^{\frac{n}{2}}\approx\left(\frac{100}{80}\right)^{\frac{n}{2}}=\left(\frac{5}{4}\right)^{\frac{n}{2}}=\left(\frac{125}{64}\right)^{\frac{n}{6}}\approx\left(\frac{128}{64}\right)^{\frac{n}{6}}=2^{\frac{n}{6}}$ .

Then we have $\frac{m}{6}\ge20$ , so we can estimate that the smallest value $m$ satisfying the condition is $m=120$ . Since this solution is well inside one of the answer choices’ bounds, we can assume this is a good enough estimate, so the answer is $\boxed{\textbf{(C) } [81,242]}$ .

~eevee9406

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

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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