2019 AMC 10B Problems/Problem 24
- 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 from and then prove 's are decreasing by Now we need to estimate the value of by since 's are decreasing, are also decreasing, so we have and which leads to The problem requires us to find the value of such that using natural logarithm, we need and , or
As estimations, and , we can estimate that Choose
Solution 2
Making the reasonable assumption that approaches , we can translate down by to obtain a more simple sequence that should approach .
Substitution of for and for in the definition of leads to
The ratio of consecutive terms is thus always positive and less than 1 (because is positive). This means that the largest possible value for is 1 and that no value of can be less than or equal to 0.
Plugging the extrema of back into the ratio shows that for all .
For , we can bound by applying this rule recursively :
Therefore, is always less than when
and is never less than when
The first integer such that must therefore lie in the interval
Both of these can be quickly estimated at c. , so the answer must be . (actual values are and )
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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