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== | ==Solution== | ||
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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 = | + | <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 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.