Difference between revisions of "2019 AMC 10B Problems/Problem 24"

Revision as of 18:48, 18 February 2019

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 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 $\boxed{\textbf{(C) } [81,242]}$ .

2019 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

2019 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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

@@ Line 9: / Line 9: @@
 == Solution 1 ==
-We first prove that <math>x_n > 4</math> for all <math>n \ge 0</math>, by induction. This follows from
+We first prove that <math>x_n > 4</math> for all <math>n \ge 0</math>, by induction. Observe that
 <cmath>
 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}
 </cmath>
+so (since <math>x_n</math> is clearly positive for all <math>n</math>, from the initial definition), <math>x_{n+1} > 4</math> if and only if <math>x_{n} > 4</math>.
 We similarly prove that <math>x_n</math> is decreasing, since
 <cmath>

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Difference between revisions of "2019 AMC 10B Problems/Problem 24"

Revision as of 18:48, 18 February 2019

Problem

Solution 1

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