2018 AIME II Problems/Problem 2
Problem
Let , , and , and for define recursively to be the remainder when ( ) is divided by . Find .
Solution 1
When given a sequence problem, one good thing to do is to check if the sequence repeats itself or if there is a pattern.
After computing more values of the sequence, it can be observed that the sequence repeats itself every 10 terms starting at .
, , , , , , , , , , , , ,
We can simplify the expression we need to solve to .
Our answer is .
Solution 2 (Overkill)
Notice that the characteristic polynomial of this is
Then since is a root, using Vieta's formula, the other two roots satisfy and .
Let and .
We have so . We found that the three roots of the characteristic polynomial are .
Now we want to express in an explicit form as .
Plugging in we get
and
so and
Hence,
Therefore
And the answer is
See Also
2018 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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