2019 AMC 10A Problems/Problem 15
- The following problem is from both the 2019 AMC 10A #15 and 2019 AMC 12A #9, so both problems redirect to this page.
Contents
Problem
A sequence of numbers is defined recursively by , , and for all Then can be written as , where and are relatively prime positive integers. What is
Solution 1 (Induction)
Using the recursive formula, we find , , and so on. It appears that , for all . Setting , we find , so the answer is .
To prove this formula, we use induction. We are given that and , which satisfy our formula. Now assume the formula holds true for all for some positive integer . By our assumption, and . Using the recursive formula, so our induction is complete.
Solution 2
We have , in other words, . So is an arithmetic sequence with step size , which means . Since the numerator and the denominator are relatively prime, the answer is .
-eric2020 (modified by Dolphindesigner)
Solution 3
It seems reasonable to transform the equation into something else. Let , , and . Therefore, we have Thus, is the harmonic mean of and . This implies is a harmonic sequence or equivalently is arithmetic. Now, we have , , , and so on. Since the common difference is , we can express explicitly as . This gives which implies . ~jakeg314
Solution 4 (Arithmetic Sequence)
Notice thatTherefore,Therefore, the sequence is an arithmetic sequence. Notice that the common difference of is and thereforeTherefore, we see that so that
~Professor-Mom
Note: This is similar to solutions #2 and #3, although you can notice that in #2's case the new sequence actually forms an arithmetic sequence.
Solution 5 (Characteristic Equation - Overkill but Generic)
We have ,
let , then
, this is 2nd order linear homogeneous recurrence sequence,
the characteristic equation for this is , which has double root x=1
so
plug in
we solve , so
so .
Since the numerator and the denominator are relatively prime, the answer is .
- note: characteristic equation is overkill for this simple one but is more generic solution for other parameter values.
See Also
2019 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 8 |
Followed by Problem 10 |
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.