Difference between revisions of "2019 AMC 10B Problems/Problem 24"
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Putting both sides to the 40th power, we get <math>2^{-20} < (y_3)^{40}</math> | Putting both sides to the 40th power, we get <math>2^{-20} < (y_3)^{40}</math> | ||
− | But y_3 = (y_0)^3, so <math>2^{-20} < (y_0)^{120}</math> and therefore, <math>2^{-20} < y_{120}</math>. | + | But <math>y_3 = (y_0)^3</math>, so <math>2^{-20} < (y_0)^{120}</math> and therefore, <math>2^{-20} < y_{120}</math>. |
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>. |
Revision as of 06:53, 24 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.
Contents
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 by mathsuper(丹神)
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 .
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.