Difference between revisions of "2019 AMC 10A Problems/Problem 15"
m (→Solution 2) |
Pi is 3.14 (talk | contribs) (→Video Solution) |
||
(24 intermediate revisions by 12 users not shown) | |||
Line 4: | Line 4: | ||
A sequence of numbers is defined recursively by <math>a_1 = 1</math>, <math>a_2 = \frac{3}{7}</math>, and | A sequence of numbers is defined recursively by <math>a_1 = 1</math>, <math>a_2 = \frac{3}{7}</math>, and | ||
− | <cmath>a_n=\frac{a_{n-2} \cdot a_{n-1}}{2a_{n-2} - a_{n-1}}</cmath>for all <math>n \geq 3</math> Then <math>a_{2019}</math> can be written as <math>\frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive | + | <cmath>a_n=\frac{a_{n-2} \cdot a_{n-1}}{2a_{n-2} - a_{n-1}}</cmath>for all <math>n \geq 3</math> Then <math>a_{2019}</math> can be written as <math>\frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. What is <math>p+q ?</math> |
<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> | ||
− | ==Solution 1== | + | ==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)== | ||
Using the recursive formula, we find <math>a_3=\frac{3}{11}</math>, <math>a_4=\frac{3}{15}</math>, and so on. It appears that <math>a_n=\frac{3}{4n-1}</math>, for all <math>n</math>. Setting <math>n=2019</math>, we find <math>a_{2019}=\frac{3}{8075}</math>, so the answer is <math>\boxed{\textbf{(E) }8078}</math>. | Using the recursive formula, we find <math>a_3=\frac{3}{11}</math>, <math>a_4=\frac{3}{15}</math>, and so on. It appears that <math>a_n=\frac{3}{4n-1}</math>, for all <math>n</math>. Setting <math>n=2019</math>, we find <math>a_{2019}=\frac{3}{8075}</math>, so the answer is <math>\boxed{\textbf{(E) }8078}</math>. | ||
− | + | To prove this formula, we use induction. We are given that <math>a_1=1</math> and <math>a_2=\frac{3}{7}</math>, which satisfy our formula. Now assume the formula holds true for all <math>n\le m</math> for some positive integer <math>m</math>. By our assumption, <math>a_{m-1}=\frac{3}{4m-5}</math> and <math>a_m=\frac{3}{4m-1}</math>. Using the recursive formula, <cmath>a_{m+1}=\frac{a_{m-1}\cdot a_m}{2a_{m-1}-a_m}=\frac{\frac{3}{4m-5}\cdot\frac{3}{4m-1}}{2\cdot\frac{3}{4m-5}-\frac{3}{4m-1}}=\frac{\left(\frac{3}{4m-5}\cdot\frac{3}{4m-1}\right)(4m-5)(4m-1)}{\left(2\cdot\frac{3}{4m-5}-\frac{3}{4m-1}\right)(4m-5)(4m-1)}=\frac{9}{6(4m-1)-3(4m-5)}=\frac{3}{4(m+1)-1},</cmath> | |
− | To prove this formula, we use induction. We are given that <math>a_1=1</math> and <math>a_2=\frac{3}{7}</math>, which satisfy our formula. Now assume the formula holds true for all <math>n\le m</math> for some positive integer <math>m</math>. By our assumption, <math>a_{m-1}=\frac{3}{4m-5}</math> and <math>a_m=\frac{3}{4m-1}</math>. Using the recursive formula, <cmath>a_{m+1}=\frac{a_{m-1}\cdot a_m}{2a_{m-1}-a_m}=\frac{\frac{3}{4m-5}\cdot\frac{3}{4m-1}}{2\cdot\frac{3}{4m-5}-\frac{3}{4m-1}}=\frac{(\frac{3}{4m-5}\cdot\frac{3}{4m-1})(4m-5)(4m-1)}{(2\cdot\frac{3}{4m-5}-\frac{3}{4m-1})(4m-5)(4m-1)}=\frac{9}{6(4m-1)-3(4m-5)}=\frac{3}{4(m+1)-1},</cmath> | ||
so our induction is complete. | so our induction is complete. | ||
==Solution 2== | ==Solution 2== | ||
− | Since we are finding the sum of the numerator and the denominator, consider the | + | Since we are interested in finding the sum of the numerator and the denominator, consider the sequence defined by <math>b_n = \frac{1}{a_n}</math>. |
+ | |||
+ | We have <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>, so in other words, <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 <math>b_n=(n-1) \cdot b_2 - (n-2) \cdot b_1</math>. | ||
+ | |||
+ | By plugging in 2019, we thus find <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>\boxed{\textbf{(E) } 8078}</math>. | ||
+ | |||
+ | -eric2020 | ||
+ | |||
+ | ==Solution 3== | ||
− | <math> | + | It seems reasonable to transform the equation into something else. Let <math>a_{n}=x</math>, <math>a_{n-1}=y</math>, and <math>a_{n-2}=z</math>. Therefore, we have <cmath>x=\frac{zy}{2z-y}</cmath> |
+ | <cmath>2xz-xy=zy</cmath> | ||
+ | <cmath>2xz=y(x+z)</cmath> | ||
+ | <cmath>y=\frac{2xz}{x+z}</cmath> | ||
+ | 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 | ||
− | <math> | + | ==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.