Difference between revisions of "2020 AIME II Problems/Problem 6"
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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 | + | 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== | ==Video Solution== |
Latest revision as of 20:10, 7 May 2021
Contents
Problem
Define a sequence recursively by , , andfor all . Then can be expressed as , where and are relatively prime positive integers. Find .
Solution
Let . Then, we have where and . By substitution, we find , , , , and . So has a period of . Thus . So, . ~mn28407
Solution 2 (Official MAA)
More generally, let the first two terms be and and replace and in the recursive formula by and , respectively. Then some algebraic calculation shows that so the sequence is periodic with period . Therefore The requested sum is .
Solution 3
Let us examine the first few terms of this sequence and see if we can find a pattern. We are obviously given and , so now we are able to determine the numerical value of using this information: Now using this information, as well as the previous information, we are able to determine the value of : Now using this information, as well as the previous information, we are able to determine the value of : Now using this information, as well as the previous information, we are able to determine the value of :
Alas, we have figured this sequence is period 5! Thus, let us take , which is , and therefore . According to the original problem statement, our answer is essentially . ~ nikenissan
Video Solution
https://youtu.be/_JTWJxbDC1A ~ CNCM
Video Solution 2
~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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