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 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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