Difference between revisions of "2019 AMC 10B Problems/Problem 24"
Sevenoptimus (talk | contribs) (Added more detail to the solution) |
Sevenoptimus (talk | contribs) m (Corrected a minor issue) |
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\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 | \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 thus reduces to finding the value of <math>n</math> such that | + | The problem thus reduces to finding the least value of <math>n</math> such that |
<cmath> | <cmath> | ||
\left(\frac{9}{10}\right)^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 } |
Revision as of 00:17, 19 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 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
.
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.