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Difference between revisions of "2020 AIME II Problems/Problem 6"
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<cmath>\implies t_5 = \frac{5(\frac{101}{525}) + 1}{25(\frac{103}{26250})}</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{101}{105} + 1}{\frac{103}{1050}}</cmath>
−
<cmath>\implies t_5 = \frac{\frac{206}{105}}{\frac{103}{1050}} \implies t_6 = 21</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
 
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

Revision as of 19:10, 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 (Problems • Answer Key • Resources)
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

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