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.
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
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.
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 .
|2019 AMC 10A (Problems • Answer Key • Resources)|
|All AMC 10 Problems and Solutions|
|2019 AMC 12A (Problems • Answer Key • Resources)|
|All AMC 12 Problems and Solutions|