Difference between revisions of "2019 AMC 10B Problems/Problem 24"
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| − | 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>\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 
and ![\[x_{n+1}=\frac{x_n^2+5x_n+4}{x_n+6}\]](http://latex.artofproblemsolving.com/9/3/4/934a32343352988b8d8328a7ef93c7553a1ce642.png)
for all nonnegative integers 
Let 
be the least positive integer such that
![\[x_m\leq 4+\frac{1}{2^{20}}.\]](http://latex.artofproblemsolving.com/e/8/8/e889e3b4b1d11c0fd3ce0eda55858624351b69fd.png)
In which of the following intervals does lie?
![$\textbf{(A) } [9,26] \qquad\textbf{(B) } [27,80] \qquad\textbf{(C) } [81,242]\qquad\textbf{(D) } [243,728] \qquad\textbf{(E) } [729,\infty)$](http://latex.artofproblemsolving.com/b/4/2/b4200e832f7695e2ceefdcf8621675991b8b23cb.png)
Solution
See Also
| 2019 AMC 12B (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 12 Problems and Solutions | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
