Difference between revisions of "2020 AIME II Problems/Problem 6"
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<cmath>\implies t_5 = \frac{5t_{4}+1}{25t_{3}}</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{5(\frac{103}{26250}) + 1}{25(\frac{53}{250})}</cmath> | ||
− | <cmath>\implies t_5 = \frac{\frac{ | + | <cmath>\implies t_5 = \frac{\frac{103}{5250} + 1}{\frac{53}{10}}</cmath> |
<cmath>\implies t_5 = \frac{\frac{5353}{5250}}{\frac{53}{10}} \implies t_5 = \frac{101}{525}</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>: | Now using this information, as well as the previous information, we are able to determine the value of <math>t_6</math>: |
Revision as of 01:37, 30 January 2022
Contents
[hide]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.