Difference between revisions of "2019 AMC 10B Problems/Problem 24"
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Solution by mathsuper(丹神) | Solution by mathsuper(丹神) | ||
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+ | ==Solution 2== | ||
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+ | The condition where <math>x_m\leq 4+\frac{1}{2^{20}}</math> gives the motivation to make a substitution to change the equilibrium from <math>4</math> to <math>0</math>. We can substitute <math>x_n = y_n + 4</math> to achieve that. | ||
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+ | Now, we need to find the smallest value of <math>m</math> such that <math>y_m\leq \frac{1}{2^{20}}</math> given that <math>y_0 = 1</math> and the recursion <math>y_{n+1} = \frac{y_n^2 + 9y_n}{y_n + 10}</math>. | ||
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+ | Using wishful thinking, we can simplify the recursion as follows: | ||
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+ | <math>y_{n+1} = \frac{y_n^2 + 9y_n + y_n - y_n}{y_n + 10}</math> | ||
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+ | <math>y_{n+1} = \frac{y_n(y_n + 10) - y_n}{y_n + 10}</math> | ||
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+ | <math>y_{n+1} = y_n - \frac{y_n}{y_n + 10}</math> | ||
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+ | <math>y_{n+1} = y_n(1 - \frac{1}{y_n + 10})</math> | ||
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+ | The recursion looks like a geometric sequence with the ratio changing slightly after each term. Notice from the recursion that the <math>y_n</math> sequence is strictly decreasing, so all the terms after <math>y_0</math> will be less than 1. Also, notice that all the terms in sequence will be positive. Both of these can be proven by induction. | ||
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+ | With both of those observations in mind, <math>\frac{9}{10} < 1 - \frac{1}{y_n + 10} \leq \frac{10}{11}</math>. Combining this with the fact that the recursion resembles a geometric sequence, we conclude that <math>(\frac{9}{10})^2 < y_n \leq (\frac{10}{11})^2</math>. | ||
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+ | Note: Solution is incomplete | ||
==See Also== | ==See Also== |
Revision as of 19:48, 20 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
.
Note: Solution is incomplete
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.