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

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<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>
 +
 +
We first prove that <math>x_n > 4</math> for all <math>n \ge 0</math> by induction from
 +
<cmath>
 +
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}
 +
</cmath>
 +
and then prove <math>x_n</math>'s are decreasing by
 +
<cmath>
 +
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
 +
</cmath>
 +
Now we need to estimate the value of <math>x_{n+1}-4</math> by
 +
<cmath>
 +
x_{n+1} - 4 = (x_n-4)\cdot\frac{x_n + 5}{x_n+6}
 +
</cmath>
 +
since <math>x_n</math>'s are decreasing, <math>\frac{x_n + 5}{x_n+6}</math> are also decreasing, so we have
 +
<cmath>
 +
\frac{9}{10} < \frac{x_n + 5}{x_n+6} \le \frac{10}{11}
 +
</cmath>
 +
and
 +
<cmath>
 +
\frac{9}{10}(x_n-4) < x_{n+1} - 4 \le \frac{10}{11}(x_n-4)
 +
</cmath>
 +
which leads to
 +
<cmath>
 +
(\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
 +
</cmath>
 +
The problem requires us to find the value of <math>n</math> such that
 +
<cmath>
 +
(\frac{9}{10})^n < x_{n} - 4 \le \frac{1}{2^{20}}
 +
</cmath>
 +
and
 +
<cmath>
 +
(\frac{10}{11})^{n-1} > x_{n-1} - 4 > \frac{1}{2^{20}}
 +
</cmath>
 +
using natural logarithm, we need
 +
<math>n \ln \frac{9}{10} < -20 \ln 2</math> and <math>(n-1)\ln \frac{10}{11} > -20 \ln 2</math>, or
 +
 +
<cmath>
 +
n > \frac{20\ln 2}{\ln\frac{10}{9}}
 +
</cmath>
 +
and
 +
<cmath>
 +
n-1 < \frac{20\ln 2}{\ln\frac{11}{10}}
 +
</cmath>
 +
 +
As estimations, <math>\ln\frac{10}{9} \approx 1/9</math> and <math>\ln\frac{11}{10} \approx 1/10</math>, <math>\ln 2\approx 0.7</math>
 +
we can estimate that
 +
<cmath>
 +
            126 < n < 141
 +
</cmath>
 +
Choose C
  
 
==See Also==
 
==See Also==

Revision as of 20:37, 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 $x_0=5$ 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?

$\textbf{(A) } [9,26] \qquad\textbf{(B) } [27,80] \qquad\textbf{(C) } [81,242]\qquad\textbf{(D) } [243,728] \qquad\textbf{(E) } [729,\infty)$

We first prove that $x_n > 4$ for all $n \ge 0$ by induction from \[x_{n+1} - 4 = \frac{x_n^2 + 5x_n + 4 - 4(x_n+6)}{x_n+6} = \frac{(x_n - 4)(x_n+5)}{x_n+6}\] and then prove $x_n$'s are decreasing by \[x_{n+1} - x_n = \frac{x_n^2 + 5x_n + 4 - x_n(x_n+6)}{x_n+6} = \frac{4-x_n}{x_n+6} < 0\] Now we need to estimate the value of $x_{n+1}-4$ by \[x_{n+1} - 4 = (x_n-4)\cdot\frac{x_n + 5}{x_n+6}\] since $x_n$'s are decreasing, $\frac{x_n + 5}{x_n+6}$ are also decreasing, so we have \[\frac{9}{10} < \frac{x_n + 5}{x_n+6} \le \frac{10}{11}\] and \[\frac{9}{10}(x_n-4) < x_{n+1} - 4 \le \frac{10}{11}(x_n-4)\] which leads to \[(\frac{9}{10})^n = (\frac{9}{10})^n (x_0-4) < x_{n} - 4 \le (\frac{10}{11})^n (x_0-4) = (\frac{10}{11})^n\] The problem requires us to find the value of $n$ such that \[(\frac{9}{10})^n < x_{n} - 4 \le \frac{1}{2^{20}}\] and \[(\frac{10}{11})^{n-1} > x_{n-1} - 4 > \frac{1}{2^{20}}\] using natural logarithm, we need $n \ln \frac{9}{10} < -20 \ln 2$ and $(n-1)\ln \frac{10}{11} > -20 \ln 2$, or

\[n > \frac{20\ln 2}{\ln\frac{10}{9}}\] and \[n-1 < \frac{20\ln 2}{\ln\frac{11}{10}}\]

As estimations, $\ln\frac{10}{9} \approx 1/9$ and $\ln\frac{11}{10} \approx 1/10$, $\ln 2\approx 0.7$ we can estimate that \[126 < n < 141\] Choose C

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

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