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

(Put in the problem...)
Line 1: Line 1:
Define a sequence recursively by <math>x_0=5</math> and
+
==Problem==
<cmath>x_{n+1}=\frac{x_n^2+5x_n+4}{x_n+6}</cmath>for all nonnegative integers <math>n.</math> Let <math>m</math> be the least positive integer such that
+
 
 +
Define a sequence recursively by <math>x_0=5</math> and <cmath>x_{n+1}=\frac{x_n^2+5x_n+4}{x_n+6}</cmath> for all nonnegative integers <math>n.</math> Let <math>m</math> be the least positive integer such that
 
<cmath>x_m\leq 4+\frac{1}{2^{20}}.</cmath>In which of the following intervals does <math>m</math> lie?
 
<cmath>x_m\leq 4+\frac{1}{2^{20}}.</cmath>In which of the following intervals does <math>m</math> lie?
  
<math>\textbf{(A) } [9,26] \qquad\textbf{(B) } [27,80] \qquad\textbf{(C) } [81,242]\qquad\textbf{(D) } [243,728] \qquad\textbf{(E) } [729,\infty]</math>
+
<math>\textbf{(A) } [9,26] \qquad\textbf{(B) } [27,80] \qquad\textbf{(C) } [81,242]\qquad\textbf{(D) } [243,728] \qquad\textbf{(E) } [729,\infty)</math>
 +
 
 +
==Solution==
 +
 
 +
 
 +
 
 +
==See Also==
 +
{{AMC12 box|year=2019|ab=B|num-b=23|num-a=25}}
 +
{{MAA Notice}}

Revision as of 17:04, 14 February 2019

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

See Also

2019 AMC 12B (ProblemsAnswer KeyResources)
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 12 Problems and Solutions

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