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

(Created page with "This page will open in around 1 hour")
 
m (Solution 3)
 
(7 intermediate revisions by 6 users not shown)
Line 1: Line 1:
This page will open in around 1 hour
+
==Problem==
 +
 
 +
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 expressed 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
 +
 
 +
==Solution 2 (Official MAA)==
 +
More generally, let the first two terms be <math>a</math> and <math>b</math> and replace <math>5</math> and <math>25</math> in the recursive formula by <math>k</math> and <math>k^2</math>, respectively. Then some algebraic calculation shows that
 +
<cmath>t_3 = \frac{b\,k+1}{a\, k^2},~~t_4 = \frac{a\, k + b\,k+1}{a\,b\, k^3},~~
 +
t_5 = \frac{a\,k+1}{b\, k^2},~~ t_6 = a, \text{~ and ~}t_7 =b,</cmath>so the sequence is periodic with period <math>5</math>. Therefore
 +
<cmath>t_{2020} = t_{5} =  \frac{20\cdot 5 +1}{21\cdot  25} = \frac{101}{525}.</cmath>The requested sum is <math>101+525=626</math>.
 +
 
 +
==Solution 3==
 +
Let us examine the first few terms of this sequence and see if we can find a pattern. We are obviously given <math>t_1 = 20</math> and <math>t_2 = 21</math>, so now we are able to determine the numerical value of <math>t_3</math> using this information:
 +
<cmath>t_3 = \frac{5t_{3-1}+1}{25t_{3-2}}</cmath>
 +
<cmath>\implies t_3 = \frac{5t_{2}+1}{25t_{1}}</cmath>
 +
<cmath>\implies t_3 = \frac{5(21) + 1}{25(20)}</cmath>
 +
<cmath>\implies t_3 = \frac{105 + 1}{500}</cmath>
 +
<cmath>\implies t_3 = \frac{106}{500} \implies t_3 = \frac{53}{250}</cmath>
 +
Now using this information, as well as the previous information, we are able to determine the value of <math>t_4</math>:
 +
<cmath>t_4 = \frac{5t_{4-1}+1}{25t_{4-2}}</cmath>
 +
<cmath>\implies t_4 = \frac{5t_{3}+1}{25t_{2}}</cmath>
 +
<cmath>\implies t_4 = \frac{5(\frac{53}{250}) + 1}{25(21)}</cmath>
 +
<cmath>\implies t_4 = \frac{\frac{53}{50} + 1}{525}</cmath>
 +
<cmath>\implies t_4 = \frac{\frac{103}{50}}{525} \implies t_4 = \frac{103}{26250}</cmath>
 +
Now using this information, as well as the previous information, we are able to determine the value of <math>t_5</math>:
 +
<cmath>t_5 = \frac{5t_{5-1}+1}{25t_{5-2}}</cmath>
 +
<cmath>\implies t_5 = \frac{5t_{4}+1}{25t_{3}}</cmath>
 +
<cmath>\implies t_5 = \frac{5(\frac{103}{26250}) + 1}{25(\frac{53}{250})}</cmath>
 +
<cmath>\implies t_5 = \frac{\frac{53}{5250} + 1}{\frac{53}{10}}</cmath>
 +
<cmath>\implies t_5 = \frac{\frac{5353}{5250}}{\frac{53}{10}} \implies t_5 = \frac{101}{525}</cmath>
 +
Now using this information, as well as the previous information, we are able to determine the value of <math>t_6</math>:
 +
<cmath>t_6 = \frac{5t_{6-1}+1}{25t_{6-2}}</cmath>
 +
<cmath>\implies t_6 = \frac{5t_{5}+1}{25t_{4}}</cmath>
 +
<cmath>\implies t_5 = \frac{5(\frac{101}{525}) + 1}{25(\frac{103}{26250})}</cmath>
 +
<cmath>\implies t_5 = \frac{\frac{101}{105} + 1}{\frac{103}{1050}}</cmath>
 +
<cmath>\implies t_5 = \frac{\frac{206}{105}}{\frac{103}{1050}} \implies t_6 = 20</cmath>
 +
 
 +
Alas, we have figured this sequence is period 5! Thus, let us take <math>2020 \pmod 5</math>, which is <math>5</math>, and therefore <math>t_{2020} = t_5 = \frac{101}{525}</math>. According to the original problem statement, our answer is essentially <math>\boxed{626}</math>. ~ nikenissan
 +
 
 +
==Video Solution==
 +
https://youtu.be/_JTWJxbDC1A ~ CNCM
 +
 
 +
==Video Solution 2==
 +
https://youtu.be/__B3pJMpfSk
 +
 
 +
~IceMatrix
 +
==See Also==
 +
{{AIME box|year=2020|n=II|num-b=5|num-a=7}}
 +
[[Category:Intermediate Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 20:10, 7 May 2021

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 expressed 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

Solution 2 (Official MAA)

More generally, let the first two terms be $a$ and $b$ and replace $5$ and $25$ in the recursive formula by $k$ and $k^2$, respectively. Then some algebraic calculation shows that \[t_3 = \frac{b\,k+1}{a\, k^2},~~t_4 = \frac{a\, k + b\,k+1}{a\,b\, k^3},~~ t_5 = \frac{a\,k+1}{b\, k^2},~~ t_6 = a, \text{~ and ~}t_7 =b,\]so the sequence is periodic with period $5$. Therefore \[t_{2020} = t_{5} =  \frac{20\cdot 5 +1}{21\cdot  25} = \frac{101}{525}.\]The requested sum is $101+525=626$.

Solution 3

Let us examine the first few terms of this sequence and see if we can find a pattern. We are obviously given $t_1 = 20$ and $t_2 = 21$, so now we are able to determine the numerical value of $t_3$ using this information: \[t_3 = \frac{5t_{3-1}+1}{25t_{3-2}}\] \[\implies t_3 = \frac{5t_{2}+1}{25t_{1}}\] \[\implies t_3 = \frac{5(21) + 1}{25(20)}\] \[\implies t_3 = \frac{105 + 1}{500}\] \[\implies t_3 = \frac{106}{500} \implies t_3 = \frac{53}{250}\] Now using this information, as well as the previous information, we are able to determine the value of $t_4$: \[t_4 = \frac{5t_{4-1}+1}{25t_{4-2}}\] \[\implies t_4 = \frac{5t_{3}+1}{25t_{2}}\] \[\implies t_4 = \frac{5(\frac{53}{250}) + 1}{25(21)}\] \[\implies t_4 = \frac{\frac{53}{50} + 1}{525}\] \[\implies t_4 = \frac{\frac{103}{50}}{525} \implies t_4 = \frac{103}{26250}\] Now using this information, as well as the previous information, we are able to determine the value of $t_5$: \[t_5 = \frac{5t_{5-1}+1}{25t_{5-2}}\] \[\implies t_5 = \frac{5t_{4}+1}{25t_{3}}\] \[\implies t_5 = \frac{5(\frac{103}{26250}) + 1}{25(\frac{53}{250})}\] \[\implies t_5 = \frac{\frac{53}{5250} + 1}{\frac{53}{10}}\] \[\implies t_5 = \frac{\frac{5353}{5250}}{\frac{53}{10}} \implies t_5 = \frac{101}{525}\] Now using this information, as well as the previous information, we are able to determine the value of $t_6$: \[t_6 = \frac{5t_{6-1}+1}{25t_{6-2}}\] \[\implies t_6 = \frac{5t_{5}+1}{25t_{4}}\] \[\implies t_5 = \frac{5(\frac{101}{525}) + 1}{25(\frac{103}{26250})}\] \[\implies t_5 = \frac{\frac{101}{105} + 1}{\frac{103}{1050}}\] \[\implies t_5 = \frac{\frac{206}{105}}{\frac{103}{1050}} \implies t_6 = 20\]

Alas, we have figured this sequence is period 5! Thus, let us take $2020 \pmod 5$, which is $5$, and therefore $t_{2020} = t_5 = \frac{101}{525}$. According to the original problem statement, our answer is essentially $\boxed{626}$. ~ nikenissan

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