Difference between revisions of "2018 AIME II Problems/Problem 2"
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==Problem== | ==Problem== | ||
− | 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} | + | 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} \cdot a_{2020} \cdot a_{2022}</math>. |
==Solution 1== | ==Solution 1== |
Latest revision as of 16:03, 12 January 2022
Contents
[hide]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 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.