Difference between revisions of "2019 AMC 10A Problems/Problem 15"

m (Improved formatting and clarity)
m (Solution 4 (Not rigorous at all))
(19 intermediate revisions by 8 users not shown)
Line 8: Line 8:
 
<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
 +
 
 +
 
 +
 
 +
==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>.
Line 24: Line 31:
  
 
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>.
 
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==
 +
 +
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
 +
 +
==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==

Revision as of 14:40, 27 December 2020

The following problem is from both the 2019 AMC 10A #15 and 2019 AMC 12A #9, so both problems redirect to this page.

Problem

A sequence of numbers is defined recursively by $a_1 = 1$, $a_2 = \frac{3}{7}$, and \[a_n=\frac{a_{n-2} \cdot a_{n-1}}{2a_{n-2} - a_{n-1}}\]for all $n \geq 3$ Then $a_{2019}$ can be written as $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. What is $p+q ?$

$\textbf{(A) } 2020 \qquad\textbf{(B) } 4039 \qquad\textbf{(C) } 6057 \qquad\textbf{(D) } 6061 \qquad\textbf{(E) } 8078$

Video Solution

https://youtu.be/h9rgKc_YVQ0

Education, the Study of Everything


Solution 1 (Induction)

Using the recursive formula, we find $a_3=\frac{3}{11}$, $a_4=\frac{3}{15}$, and so on. It appears that $a_n=\frac{3}{4n-1}$, for all $n$. Setting $n=2019$, we find $a_{2019}=\frac{3}{8075}$, so the answer is $\boxed{\textbf{(E) }8078}$.

To prove this formula, we use induction. We are given that $a_1=1$ and $a_2=\frac{3}{7}$, which satisfy our formula. Now assume the formula holds true for all $n\le m$ for some positive integer $m$. By our assumption, $a_{m-1}=\frac{3}{4m-5}$ and $a_m=\frac{3}{4m-1}$. Using the recursive formula, \[a_{m+1}=\frac{a_{m-1}\cdot a_m}{2a_{m-1}-a_m}=\frac{\frac{3}{4m-5}\cdot\frac{3}{4m-1}}{2\cdot\frac{3}{4m-5}-\frac{3}{4m-1}}=\frac{\left(\frac{3}{4m-5}\cdot\frac{3}{4m-1}\right)(4m-5)(4m-1)}{\left(2\cdot\frac{3}{4m-5}-\frac{3}{4m-1}\right)(4m-5)(4m-1)}=\frac{9}{6(4m-1)-3(4m-5)}=\frac{3}{4(m+1)-1},\] 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 $b_n = \frac{1}{a_n}$.

We have $\frac{1}{a_n} = \frac{2a_{n-2}-a_{n-1}}{a_{n-2} \cdot a_{n-1}}=\frac{2}{a_{n-1}}-\frac{1}{a_{n-2}}$, so in other words, $b_n = 2b_{n-1}-b_{n-2}=3b_{n-2}-2b_{n-3}=4b_{n-3}-3b_{n-4}=\dots$.

By recursively following this pattern, we can see that $b_n=(n-1) \cdot b_2 - (n-2) \cdot b_1$.

By plugging in 2019, we thus find $b_{2019} = 2018 \cdot \frac{7}{3}-2017 = \frac{8075}{3}$. Since the numerator and the denominator are relatively prime, the answer is $\boxed{\textbf{(E) } 8078}$.

-eric2020

Solution 3

It seems reasonable to transform the equation into something else. Let $a_{n}=x$, $a_{n-1}=y$, and $a_{n-2}=z$. Therefore, we have \[x=\frac{zy}{2z-y}\] \[2xz-xy=zy\] \[2xz=y(x+z)\] \[y=\frac{2xz}{x+z}\] Thus, $y$ is the harmonic mean of $x$ and $z$. This implies $a_{n}$ is a harmonic sequence or equivalently $b_{n}=\frac{1}{a_{n}}$ is arithmetic. Now, we have $b_{1}=1$, $b_{2}=\frac{7}{3}$, $b_{3}=\frac{11}{3}$, and so on. Since the common difference is $\frac{4}{3}$, we can express $b_{n}$ explicitly as $b_{n}=\frac{4}{3}(n-1)+1$. This gives $b_{2019}=\frac{4}{3}(2019-1)+1=\frac{8075}{3}$ which implies $a_{2019}=\frac{3}{8075}=\frac{p}{q}$. $p+q=\boxed{\textbf{(E) } 8078}$ ~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: $1, \frac{3}{7}, \frac{3}{11}, \frac{1}{5}, \frac{3}{19}, \frac{3}{23} ...$

Now, we notice that the numerator seems to be in a pattern: $1, 3, 3, 1, 3, 3, 1, 3, 3...$ Then, we notice that the only time the numerator is $1$ is when the index is a multiple of $4$. Clearly, $2019$ is NOT a multiple of $4$, so the numerator is $3$. Then, using the positions of each term, we can come up with a simple formula for the denominator with $n$ as the position or index (This only applies for the numbers with numerator $3$): $3n + (n - 1)$.

Plugging $n$ in for $2019$, we get $8075$ for the denominator. Adding $3$ (The numerator) gives $\boxed{\textbf{(E) }8078}$

~EricShi1685

See Also

2019 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS