Difference between revisions of "2020 AIME II Problems/Problem 6"
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==Video Solution 2== | ==Video Solution 2== | ||
https://youtu.be/__B3pJMpfSk | https://youtu.be/__B3pJMpfSk | ||
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+ | ==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}<math>. t_4 can then be quickly calculated to around \frac{1}{250}</math>, and t_5 can be rounded to \frac{1}{5}$. t_6 turns out to be around 20, which means that there is probably a loop with period 5. The rest of the solution proceeds normally. | ||
~IceMatrix | ~IceMatrix |
Revision as of 00:00, 25 December 2022
Contents
[hide]Problem
Define a sequence recursively by , , andfor all . Then can be expressed as , where and are relatively prime positive integers. Find .
Solution
Let . Then, we have where and . By substitution, we find , , , , and . So has a period of . Thus . So, . ~mn28407
Solution 2 (Official MAA)
More generally, let the first two terms be and and replace and in the recursive formula by and , respectively. Then some algebraic calculation shows that so the sequence is periodic with period . Therefore The requested sum is .
Solution 3
Let us examine the first few terms of this sequence and see if we can find a pattern. We are obviously given and , so now we are able to determine the numerical value of using this information:
Alas, we have figured this sequence is period 5! But since , we can state that . According to the original problem statement, our answer is . ~ nikenissan
Video Solution
https://youtu.be/_JTWJxbDC1A ~ CNCM
Video Solution 2
Quick way to notice recursion loop
Round the first two values to both be 20. Then, the next element can be rounded to \frac{1}{5}, and t_5 can be rounded to \frac{1}{5}$. 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.
~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.