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
Since we are interested in finding the sum of the numerator and the denominator, consider the sequence defined by .
We have , so in other words, .
By recursively following this pattern, we can see that .
By plugging in 2019, we thus find . Since the numerator and the denominator are relatively prime, the answer is .
-eric2020
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 (Not rigorous at all)
Noticing that there is clearly a pattern, but the formula for it is hidious, we first find the first few terms of the sequence to see if there is any pattern:
Now, we notice that the numerator seems to be in a pattern: Then, we notice that the only time the numerator is is when the index is a multiple of . Clearly, is NOT a multiple of , so the numerator is . Then, using the positions of each term, we can come up with a simple formula for the denominator with as the position or index (This only applies for the numbers with numerator ): .
Plugging in for , we get for the denominator. Adding (The numerator) gives
~EricShi1685
Solution 5
The first 4 terms are: Therefore the answer is . ~Lcz
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 |
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