2018 AIME II Problems/Problem 2

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Problem

Let $a_{0} = 2$, $a_{1} = 5$, and $a_{2} = 8$, and for $n > 2$ define $a_{n}$ recursively to be the remainder when $4$($a_{n-1}$ $+$ $a_{n-2}$ $+$ $a_{n-3}$) is divided by $11$. Find $a_{2018}$$a_{2020}$$a_{2022}$.

Solution 2

Solution

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 $a_{0}$.

$a_{0} = 2$, $a_{1} = 5$, $a_{2} = 8$, $a_{3} = 5$, $a_{4} = 6$, $a_{5} = 10$, $a_{6} = 7$, $a_{7} = 4$, $a_{8} = 7$, $a_{9} = 6$, $a_{10} = 2$, $a_{11} = 5$, $a_{12} = 8$, $a_{13} = 5$

We can simplify the expression we need to solve to $a_{8}$$a_{10}$$a_{2}$.

Our answer is $7$$2$$8$ $= \boxed{112}$.

See Also

2018 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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