2018 AIME II Problems/Problem 2

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

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 $7$ • $2$ • $8$ $= \boxed{112}$ .

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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2018 AIME II Problems/Problem 2

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