Difference between revisions of "2019 AMC 10B Problems/Problem 24"
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This tells us that <math>m</math> is somewhere around 120, so our answer is <math>\boxed{\textbf{(C) } [81,242]}</math>. | This tells us that <math>m</math> is somewhere around 120, so our answer is <math>\boxed{\textbf{(C) } [81,242]}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | Since the choices are rather wide ranges, we can use approximation to make it easier. Notice that | ||
+ | <cmath>x_{n+1} - x_n = \frac{4-x_n}{x_n+6}</cmath> | ||
+ | And <math>x_0 =5</math>, we know that <math>x_n</math> is a declining sequence, and as it get close to 4 its decline will slow, never falling below 4. So we'll use 4 to approximate <math>x_n</math> in the denominator so that we have a solvable difference equation: | ||
+ | <cmath> x_{n+1} - x_n = \frac{4-x_n}{10}</cmath> | ||
+ | <cmath>x_{n+1} = \frac{9}{10}x_n + \frac{2}{5}</cmath> | ||
+ | Solve it with <math>x_0 = 5</math>, we have | ||
+ | <cmath>x_n = 4 + (\frac{9}{10})^n</cmath> | ||
+ | Now we wish to find <math>n</math> so that | ||
+ | <cmath> (\frac{9}{10})^n \approx \frac{1}{2^{20}}</cmath> | ||
+ | <cmath> n \approx \frac{\log{2^{20}}}{\log{10}-\log9} \approx \frac{20*0.3}{0.05} = 120</cmath> | ||
+ | Since 120 is safely within the range of [81,242], we have the answer. | ||
+ | <math>\boxed{\textbf{(C) } [81,242]}</math>. | ||
+ | |||
+ | -Mathdummy | ||
==Video Solution== | ==Video Solution== |
Latest revision as of 10:40, 3 December 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 for all nonnegative integers Let be the least positive integer such that In which of the following intervals does lie?
Solution 1
We first prove that for all , by induction. Observe that so (since is clearly positive for all , from the initial definition), if and only if .
We similarly prove that is decreasing, since
Now we need to estimate the value of , which we can do using the rearranged equation Since is decreasing, is clearly also decreasing, so we have and
This becomes The problem thus reduces to finding the least value of such that
Taking logarithms, we get and , i.e.
As approximations, we can use , , and . These allow us to estimate that which gives the answer as .
Solution 2
The condition where gives the motivation to make a substitution to change the equilibrium from to . We can substitute to achieve that.
Now, we need to find the smallest value of such that given that and the recursion .
Using wishful thinking, we can simplify the recursion as follows:
The recursion looks like a geometric sequence with the ratio changing slightly after each term. Notice from the recursion that the sequence is strictly decreasing, so all the terms after will be less than 1. Also, notice that all the terms in sequence will be positive. Both of these can be proven by induction.
With both of those observations in mind, . Combining this with the fact that the recursion resembles a geometric sequence, we conclude that .
is approximately equal to and the ranges that the answer choices give us are generous, so we should use either or to find a rough estimate for . is , while is close to because is , which is close to .
Therefore, we can estimate that .
Putting both sides to the 40th power, we get
But , so and therefore, .
This tells us that is somewhere around 120, so our answer is .
Solution 3
Since the choices are rather wide ranges, we can use approximation to make it easier. Notice that And , we know that is a declining sequence, and as it get close to 4 its decline will slow, never falling below 4. So we'll use 4 to approximate in the denominator so that we have a solvable difference equation: Solve it with , we have Now we wish to find so that Since 120 is safely within the range of [81,242], we have the answer. .
-Mathdummy
Video Solution
Video Solution: https://www.youtube.com/watch?v=Ok7bYOdiF6M
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.