Difference between revisions of "2020 AIME II Problems/Problem 6"

m
(Written Solution Added)
Line 3: Line 3:
 
Define a sequence recursively by <math>t_1 = 20</math>, <math>t_2 = 21</math>, and<cmath>t_n = \frac{5t_{n-1}+1}{25t_{n-2}}</cmath>for all <math>n \ge 3</math>. Then <math>t_{2020}</math> can be written as <math>\frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.
 
Define a sequence recursively by <math>t_1 = 20</math>, <math>t_2 = 21</math>, and<cmath>t_n = \frac{5t_{n-1}+1}{25t_{n-2}}</cmath>for all <math>n \ge 3</math>. Then <math>t_{2020}</math> can be written as <math>\frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.
  
 +
==Solution==
 +
Let <math>t_n=\frac{s_n}{5}</math>. Then, we have <math>s_n=\frac{s_{n-1}+1}{s_{n-2}}</math> where <math>s_1 = 100</math> and <math>s_2 = 105</math>. By substitution, we find <math>s_3 = \frac{53}{50}</math>, <math>s_4=\frac{103}{105\cdot50}</math>, <math>s_5=\frac{101}{105}</math>, <math>s_6=100</math>, and <math>s_7=105</math>. So <math>s_n</math> has a period of <math>5</math>. Thus <math>s_{2020}=s_5=\frac{101}{105}</math>. So, <math>\frac{101}{105\cdot 5}\implies 101+525=\boxed{626}</math>.
 +
~mn28407
  
 
==Video Solution==
 
==Video Solution==

Revision as of 03:13, 8 June 2020

Problem

Define a sequence recursively by $t_1 = 20$, $t_2 = 21$, and\[t_n = \frac{5t_{n-1}+1}{25t_{n-2}}\]for all $n \ge 3$. Then $t_{2020}$ can be written as $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.

Solution

Let $t_n=\frac{s_n}{5}$. Then, we have $s_n=\frac{s_{n-1}+1}{s_{n-2}}$ where $s_1 = 100$ and $s_2 = 105$. By substitution, we find $s_3 = \frac{53}{50}$, $s_4=\frac{103}{105\cdot50}$, $s_5=\frac{101}{105}$, $s_6=100$, and $s_7=105$. So $s_n$ has a period of $5$. Thus $s_{2020}=s_5=\frac{101}{105}$. So, $\frac{101}{105\cdot 5}\implies 101+525=\boxed{626}$. ~mn28407

Video Solution

https://youtu.be/_JTWJxbDC1A ~ CNCM

Video Solution 2

https://youtu.be/__B3pJMpfSk

~IceMatrix

See Also

2020 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME 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