2019 AMC 10B Problems/Problem 24: Difference between revisions
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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
for all nonnegative integers
Let
be the least positive integer such that
In which of the following intervals does
lie?
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 | |
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