Difference between revisions of "2019 AMC 10A Problems/Problem 15"
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By recursively following this pattern, we can see that <math>b_n=(n-1) \cdot b_2 - (n-1) \cdot b_1</math>. | 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: <math>b_{2019} = 2018 \cdot \frac{7}{3}-2017 = \frac{ | + | By plugging in 2019, we get: <math>b_{2019} = 2018 \cdot \frac{7}{3}-2017 = \frac{8075}{3}</math>. Since the numerator and the denominator are relatively prime, the answer is <math>8078</math>, which implies <math>\implies \boxed{E}</math> |
-eric2020 | -eric2020 |
Revision as of 23:35, 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 the numerator and the denominator are relatively prime, the answer is
, which implies
-eric2020
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.