Difference between revisions of "2019 AMC 10A Problems/Problem 15"
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<math>\textbf{(A) } 2020 \qquad\textbf{(B) } 4039 \qquad\textbf{(C) } 6057 \qquad\textbf{(D) } 6061 \qquad\textbf{(E) } 8078</math> | <math>\textbf{(A) } 2020 \qquad\textbf{(B) } 4039 \qquad\textbf{(C) } 6057 \qquad\textbf{(D) } 6061 \qquad\textbf{(E) } 8078</math> | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/h9rgKc_YVQ0 | ||
+ | |||
+ | Education, the Study of Everything | ||
+ | |||
+ | == Video Solution (Meta-Solving Technique) == | ||
+ | https://youtu.be/GmUWIXXf_uk?t=39 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==Solution 1 (Induction)== | ==Solution 1 (Induction)== | ||
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Thus, <math>y</math> is the harmonic mean of <math>x</math> and <math>z</math>. This implies <math>a_{n}</math> is a harmonic sequence or equivalently <math>b_{n}=\frac{1}{a_{n}}</math> is arithmetic. Now, we have <math>b_{1}=1</math>, <math>b_{2}=\frac{7}{3}</math>, <math>b_{3}=\frac{11}{3}</math>, and so on. Since the common difference is <math>\frac{4}{3}</math>, we can express <math>b_{n}</math> explicitly as <math>b_{n}=\frac{4}{3}(n-1)+1</math>. This gives <math>b_{2019}=\frac{4}{3}(2019-1)+1=\frac{8075}{3}</math> which implies <math>a_{2019}=\frac{3}{8075}=\frac{p}{q}</math>. <math>p+q=\boxed{\textbf{(E) } 8078}</math> | Thus, <math>y</math> is the harmonic mean of <math>x</math> and <math>z</math>. This implies <math>a_{n}</math> is a harmonic sequence or equivalently <math>b_{n}=\frac{1}{a_{n}}</math> is arithmetic. Now, we have <math>b_{1}=1</math>, <math>b_{2}=\frac{7}{3}</math>, <math>b_{3}=\frac{11}{3}</math>, and so on. Since the common difference is <math>\frac{4}{3}</math>, we can express <math>b_{n}</math> explicitly as <math>b_{n}=\frac{4}{3}(n-1)+1</math>. This gives <math>b_{2019}=\frac{4}{3}(2019-1)+1=\frac{8075}{3}</math> which implies <math>a_{2019}=\frac{3}{8075}=\frac{p}{q}</math>. <math>p+q=\boxed{\textbf{(E) } 8078}</math> | ||
~jakeg314 | ~jakeg314 | ||
+ | |||
+ | ==Solution 4 (Not rigorous at all)== | ||
+ | Noticing that there is clearly a pattern, but the formula for it is hideous, we first find the first few terms of the sequence to see if there is any pattern: <math>1, \frac{3}{7}, \frac{3}{11}, \frac{1}{5}, \frac{3}{19}, \frac{3}{23} ...</math> | ||
+ | |||
+ | Now, we notice that the numerator seems to be in a pattern: <math>1, 3, 3, 1, 3, 3, 1, 3, 3...</math> Then, we notice that the only time the numerator is <math>1</math> is when the index is a multiple of <math>4</math>. Clearly, <math>2019</math> is NOT a multiple of <math>4</math>, so the numerator is <math>3</math>. Then, using the positions of each term, we can come up with a simple formula for the denominator with <math>n</math> as the position or index (This only applies for the numbers with numerator <math>3</math>): <math>3n + (n - 1)</math>. | ||
+ | |||
+ | Plugging <math>n</math> in for <math>2019</math>, we get <math>8075</math> for the denominator. Adding <math>3</math> (The numerator) gives <math>\boxed{\textbf{(E) }8078}</math> | ||
+ | |||
+ | ~EricShi1685 | ||
==See Also== | ==See Also== |
Latest revision as of 22:02, 24 January 2021
- 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
Video Solution
Education, the Study of Everything
Video Solution (Meta-Solving Technique)
https://youtu.be/GmUWIXXf_uk?t=39
~ pi_is_3.14
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 hideous, 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
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.