Difference between revisions of "2018 AIME II Problems/Problem 2"
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Let <math>a_{0} = 2</math>, <math>a_{1} = 5</math>, and <math>a_{2} = 8</math>, and for <math>n > 2</math> define <math>a_{n}</math> recursively to be the remainder when <math>4</math>(<math>a_{n-1}</math> <math>+</math> <math>a_{n-2}</math> <math>+</math> <math>a_{n-3}</math>) is divided by <math>11</math>. Find <math>a_{2018}</math> • <math>a_{2020}</math> • <math>a_{2022}</math>. | Let <math>a_{0} = 2</math>, <math>a_{1} = 5</math>, and <math>a_{2} = 8</math>, and for <math>n > 2</math> define <math>a_{n}</math> recursively to be the remainder when <math>4</math>(<math>a_{n-1}</math> <math>+</math> <math>a_{n-2}</math> <math>+</math> <math>a_{n-3}</math>) is divided by <math>11</math>. Find <math>a_{2018}</math> • <math>a_{2020}</math> • <math>a_{2022}</math>. | ||
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| + | ===Solution 2=== | ||
==Solution== | ==Solution== | ||
Revision as of 16:37, 12 April 2018
Contents
Problem
Let 
, 
, and 
, and for 
define 
recursively to be the remainder when 
(
) is divided by 
. Find 
• 
• 
.
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 
.
,
,
,
,
,
,
,
,
,
,
,
,
,
We can simplify the expression we need to solve to 
• 
• 
.
Our 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 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
