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

(→‎See Also)
m (Cleaned up formatting and made the solution more readable)
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== Solution 1 ==
 
== Solution 1 ==
−
We first prove that <math>x_n > 4</math> for all <math>n \ge 0</math> by induction from  
+
We first prove that <math>x_n > 4</math> for all <math>n \ge 0</math>, by induction. This follows from
 
<cmath>
 
<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}
 
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>
 
</cmath>
−
and then prove <math>x_n</math>'s are decreasing by
+
We similarly prove that <math>x_n</math> is decreasing, since
 
<cmath>
 
<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
 
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>
 
</cmath>
−
Now we need to estimate the value of <math>x_{n+1}-4</math> by
+
 
 +
Now we need to estimate the value of <math>x_{n+1}-4</math>, which we can do using the rearranged equation
 
<cmath>
 
<cmath>
 
x_{n+1} - 4 = (x_n-4)\cdot\frac{x_n + 5}{x_n+6}  
 
x_{n+1} - 4 = (x_n-4)\cdot\frac{x_n + 5}{x_n+6}  
 
</cmath>
 
</cmath>
−
since <math>x_n</math>'s are decreasing, <math>\frac{x_n + 5}{x_n+6}</math> are also decreasing, so we have
+
Since <math>x_n</math> is decreasing, <math>\frac{x_n + 5}{x_n+6}</math> is clearly also decreasing, so we have
 
<cmath>
 
<cmath>
 
\frac{9}{10} < \frac{x_n + 5}{x_n+6} \le \frac{10}{11}
 
\frac{9}{10} < \frac{x_n + 5}{x_n+6} \le \frac{10}{11}
Line 29: Line 30:
 
\frac{9}{10}(x_n-4) < x_{n+1} - 4 \le \frac{10}{11}(x_n-4)
 
\frac{9}{10}(x_n-4) < x_{n+1} - 4 \le \frac{10}{11}(x_n-4)
 
</cmath>
 
</cmath>
−
which leads to
+
 
 +
This becomes
 
<cmath>
 
<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
+
\left(\frac{9}{10}\right)^n = \left(\frac{9}{10}\right)^n \left(x_0-4\right) < x_{n} - 4 \le \left(\frac{10}{11}\right)^n \left(x_0-4\right) = \left(\frac{10}{11}\right)^n
 
</cmath>
 
</cmath>
−
The problem requires us to find the value of <math>n</math> such that
+
The problem thus reduces to finding the value of <math>n</math> such that
 
<cmath>
 
<cmath>
−
(\frac{9}{10})^n < x_{n} - 4 \le \frac{1}{2^{20}} \text{  and  }  
+
\left(\frac{9}{10}\right)^n < x_{n} - 4 \le \frac{1}{2^{20}} \text{  and  }  
−
(\frac{10}{11})^{n-1} > x_{n-1} - 4 > \frac{1}{2^{20}}
+
\left(\frac{10}{11}\right)^{n-1} > x_{n-1} - 4 > \frac{1}{2^{20}}
 
</cmath>
 
</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
+
Taking logarithms, we get
 +
<math>n \ln \frac{9}{10} < -20 \ln 2</math> and <math>(n-1)\ln \frac{10}{11} > -20 \ln 2</math>, i.e.
  
 
<cmath>
 
<cmath>
Line 45: Line 48:
 
</cmath>
 
</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>
+
As approximations, we can use <math>\ln\frac{10}{9} \approx \frac{1}{9}</math>, <math>\ln\frac{11}{10} \approx \frac{1}{10}</math>, and <math>\ln 2\approx 0.7</math>. These allow us to estimate that
−
we can estimate that
 
 
<cmath>
 
<cmath>
 
             126 < n < 141
 
             126 < n < 141
 
</cmath>
 
</cmath>
−
Giving us <math>\boxed{\textbf{(C) } [81,242]}</math>.
+
which gives the answer as <math>\boxed{\textbf{(C) } [81,242]}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 23:58, 17 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?

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

Solution 1

We first prove that $x_n > 4$ for all $n \ge 0$, by induction. This follows from \[x_{n+1} - 4 = \frac{x_n^2 + 5x_n + 4 - 4(x_n+6)}{x_n+6} = \frac{(x_n - 4)(x_n+5)}{x_n+6}\] We similarly prove that $x_n$ is decreasing, since \[x_{n+1} - x_n = \frac{x_n^2 + 5x_n + 4 - x_n(x_n+6)}{x_n+6} = \frac{4-x_n}{x_n+6} < 0\]

Now we need to estimate the value of $x_{n+1}-4$, which we can do using the rearranged equation \[x_{n+1} - 4 = (x_n-4)\cdot\frac{x_n + 5}{x_n+6}\] Since $x_n$ is decreasing, $\frac{x_n + 5}{x_n+6}$ is clearly also decreasing, so we have \[\frac{9}{10} < \frac{x_n + 5}{x_n+6} \le \frac{10}{11}\] and \[\frac{9}{10}(x_n-4) < x_{n+1} - 4 \le \frac{10}{11}(x_n-4)\]

This becomes \[\left(\frac{9}{10}\right)^n = \left(\frac{9}{10}\right)^n \left(x_0-4\right) < x_{n} - 4 \le \left(\frac{10}{11}\right)^n \left(x_0-4\right) = \left(\frac{10}{11}\right)^n\] The problem thus reduces to finding the value of $n$ such that \[\left(\frac{9}{10}\right)^n < x_{n} - 4 \le \frac{1}{2^{20}} \text{ and } \left(\frac{10}{11}\right)^{n-1} > x_{n-1} - 4 > \frac{1}{2^{20}}\]

Taking logarithms, we get $n \ln \frac{9}{10} < -20 \ln 2$ and $(n-1)\ln \frac{10}{11} > -20 \ln 2$, i.e.

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

As approximations, we can use $\ln\frac{10}{9} \approx \frac{1}{9}$, $\ln\frac{11}{10} \approx \frac{1}{10}$, and $\ln 2\approx 0.7$. These allow us to estimate that \[126 < n < 141\] which gives the answer as $\boxed{\textbf{(C) } [81,242]}$.

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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