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

(Posted another solution)
 
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==Problem==
 
==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 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 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==
 
==Solution==
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==Solution 3==
 
==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:
 
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>t_3 = \frac{5t_{3-1}+1}{25t_{3-2}} = \frac{5t_{2}+1}{25t_{1}} = \frac{5(21) + 1}{25(20)} = \frac{105 + 1}{500}t_3 = \frac{106}{500} = \frac{53}{250}</cmath>
−
<cmath>\implies t_3 = \frac{5t_{2}+1}{25t_{1}}</cmath>
+
<cmath>t_4 = \frac{5t_{4-1}+1}{25t_{4-2}} = \frac{5t_{3}+1}{25t_{2}} = \frac{5(\frac{53}{250}) + 1}{25(21)} = \frac{\frac{53}{50} + 1}{525} = \frac{\frac{103}{50}}{525} = \frac{103}{26250}</cmath>
−
<cmath>\implies t_3 = \frac{5(21) + 1}{25(20)}</cmath>
+
<cmath>t_5 = \frac{5t_{5-1}+1}{25t_{5-2}} = \frac{5t_{4}+1}{25t_{3}} = \frac{5(\frac{103}{26250}) + 1}{25(\frac{53}{250})} = \frac{\frac{103}{5250} + 1}{\frac{53}{10}} = \frac{\frac{5353}{5250}}{\frac{53}{10}} = \frac{101}{525}</cmath>
−
<cmath>\implies t_3 = \frac{105 + 1}{500}</cmath>
+
<cmath>t_6 = \frac{5t_{6-1}+1}{25t_{6-2}} = \frac{5t_{5}+1}{25t_{4}} = \frac{5(\frac{101}{525}) + 1}{25(\frac{103}{26250})} = \frac{\frac{101}{105} + 1}{\frac{103}{1050}} = \frac{\frac{206}{105}}{\frac{103}{1050}} \implies t_6 = 20</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 = 21</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! But since <math>2020 \equiv 5 \pmod 5</math>, we can state that <math>t_{2020} = t_5 = \frac{101}{525}</math>. According to the original problem statement, our answer is <math>\boxed{626}</math>. ~ nikenissan
  
 
==Video Solution==
 
==Video Solution==
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~IceMatrix
 
~IceMatrix
 +
 +
==Quick way to notice recursion loop==
 +
Round the first two values to both be 20. Then, the next element can be rounded to <math>\frac{1}{5}</math>. <math>t_4</math> can then be quickly calculated to around <math>\frac{1}{250}</math>, and <math>t_5</math> can be rounded to <math>\frac{1}{5}</math>. <math>t_6</math> turns out to be around 20, which means that there is probably a loop with period 5. The rest of the solution proceeds normally.
 +
 
==See Also==
 
==See Also==
 
{{AIME box|year=2020|n=II|num-b=5|num-a=7}}
 
{{AIME box|year=2020|n=II|num-b=5|num-a=7}}
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 00:03, 25 December 2022

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: \[t_3 = \frac{5t_{3-1}+1}{25t_{3-2}} = \frac{5t_{2}+1}{25t_{1}} = \frac{5(21) + 1}{25(20)} = \frac{105 + 1}{500}t_3 = \frac{106}{500} = \frac{53}{250}\] \[t_4 = \frac{5t_{4-1}+1}{25t_{4-2}} = \frac{5t_{3}+1}{25t_{2}} = \frac{5(\frac{53}{250}) + 1}{25(21)} = \frac{\frac{53}{50} + 1}{525} = \frac{\frac{103}{50}}{525} = \frac{103}{26250}\] \[t_5 = \frac{5t_{5-1}+1}{25t_{5-2}} = \frac{5t_{4}+1}{25t_{3}} = \frac{5(\frac{103}{26250}) + 1}{25(\frac{53}{250})} = \frac{\frac{103}{5250} + 1}{\frac{53}{10}} = \frac{\frac{5353}{5250}}{\frac{53}{10}} = \frac{101}{525}\] \[t_6 = \frac{5t_{6-1}+1}{25t_{6-2}} = \frac{5t_{5}+1}{25t_{4}} = \frac{5(\frac{101}{525}) + 1}{25(\frac{103}{26250})} = \frac{\frac{101}{105} + 1}{\frac{103}{1050}} = \frac{\frac{206}{105}}{\frac{103}{1050}} \implies t_6 = 20\]

Alas, we have figured this sequence is period 5! But since $2020 \equiv 5 \pmod 5$, we can state that $t_{2020} = t_5 = \frac{101}{525}$. According to the original problem statement, our answer is $\boxed{626}$. ~ nikenissan

Video Solution

https://youtu.be/_JTWJxbDC1A ~ CNCM

Video Solution 2

https://youtu.be/__B3pJMpfSk

~IceMatrix

Quick way to notice recursion loop

Round the first two values to both be 20. Then, the next element can be rounded to $\frac{1}{5}$. $t_4$ can then be quickly calculated to around $\frac{1}{250}$, and $t_5$ can be rounded to $\frac{1}{5}$. $t_6$ turns out to be around 20, which means that there is probably a loop with period 5. The rest of the solution proceeds normally.

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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