Difference between revisions of "2019 AMC 10A Problems/Problem 15"
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<math>\frac{1}{a_n} = \frac{2a_{n-2}-a_{n-1}}{a_{n-2} \cdot a_{n-1}}=\frac{2}{a_{n-1}}-\frac{1}{a_{n-2}}</math> | <math>\frac{1}{a_n} = \frac{2a_{n-2}-a_{n-1}}{a_{n-2} \cdot a_{n-1}}=\frac{2}{a_{n-1}}-\frac{1}{a_{n-2}}</math> | ||
− | <math>b_n = 2b_{n-1}-b_{n-2}=3b_{n-2}-2b_{n-3} | + | <math>b_n = 2b_{n-1}-b_{n-2}=3b_{n-2}-2b_{n-3}=4b_{n-3}-3b_{n-4}=\dots</math> |
− | By recursively following this pattern, we can see that < | + | By recursively following this pattern, we can see that <math>b_n=(n-1) \cdot b_2 - (n-1) \cdot b_1</math>. |
− | By plugging in 2019, we get: < | + | By plugging in 2019, we get: <math>b_{2019} = 2018 \cdot \frac{7}{3}-2017 = \frac{8705}{3}</math>. Since they are relatively prime, the answer is <math>8708</math> which implies <math>\rightarrow \boxed{E}</math> |
==See Also== | ==See Also== |
Revision as of 17:40, 9 February 2019
- 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 inegers. What is
Solution 1
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 finding the sum of the numerator and the denominator, consider the function .
By recursively following this pattern, we can see that .
By plugging in 2019, we get: . Since they are relatively prime, the answer is which implies
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.