Difference between revisions of "2019 AMC 10B Problems/Problem 24"
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==See Also== | ==See Also== | ||
Revision as of 21:04, 14 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 
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 1
We first prove that 
for all 
by induction from
![\[x_{n+1} - 4 = \frac{x_n^2 + 5x_n + 4 - 4(x_n+6)}{x_n+6} = \frac{(x_n - 4)(x_n+5)}{x_n+6}\]](http://latex.artofproblemsolving.com/c/6/9/c692909b8b22d8ad14462307b7761004e6d2c45c.png)
and then prove 
's are decreasing by
![\[x_{n+1} - x_n = \frac{x_n^2 + 5x_n + 4 - x_n(x_n+6)}{x_n+6} = \frac{4-x_n}{x_n+6} < 0\]](http://latex.artofproblemsolving.com/4/e/0/4e047832a39ed3c356152520ca4d39a32cc56245.png)
Now we need to estimate the value of 
by
![\[x_{n+1} - 4 = (x_n-4)\cdot\frac{x_n + 5}{x_n+6}\]](http://latex.artofproblemsolving.com/6/d/e/6dea77ec20f70e23a10ebc0cc62a246de56e606f.png)
since 's are decreasing, 
are also decreasing, so we have
![\[\frac{9}{10} < \frac{x_n + 5}{x_n+6} \le \frac{10}{11}\]](http://latex.artofproblemsolving.com/f/c/c/fcc1383d3329ce9633e8b64797b4b08c835823a0.png)
and
![\[\frac{9}{10}(x_n-4) < x_{n+1} - 4 \le \frac{10}{11}(x_n-4)\]](http://latex.artofproblemsolving.com/b/f/9/bf9660c350a5f46ffe5aa358c360fd688b56db24.png)
which leads to
![\[(\frac{9}{10})^n = (\frac{9}{10})^n (x_0-4) < x_{n} - 4 \le (\frac{10}{11})^n (x_0-4) = (\frac{10}{11})^n\]](http://latex.artofproblemsolving.com/2/1/2/212a3b36b76b659cd9f2d9b4eb276483e4d3ece0.png)
The problem requires us to find the value of 
such that
![\[(\frac{9}{10})^n < x_{n} - 4 \le \frac{1}{2^{20}} \text{ and } (\frac{10}{11})^{n-1} > x_{n-1} - 4 > \frac{1}{2^{20}}\]](http://latex.artofproblemsolving.com/0/a/c/0ace8a0d4379794efd453de064305049f82b2909.png)
using natural logarithm, we need
and 
, or
![\[n > \frac{20\ln 2}{\ln\frac{10}{9}} \text{ and } n-1 < \frac{20\ln 2}{\ln\frac{11}{10}}\]](http://latex.artofproblemsolving.com/d/e/d/ded556e0aea858e38b44b191fc75ddd065f07ee4.png)
As estimations, 
and 
, 
we can estimate that
![\[126 < n < 141\]](http://latex.artofproblemsolving.com/7/2/5/7255d5386ce08cbfc583a33f37785ccf8206d63c.png)
Choose 
~Solution by mathsuper(丹神)
See Also
| 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. 
