Difference between revisions of "2018 AIME II Problems/Problem 2"
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Hence, <math>a_n\equiv 6+(6^n)+6(8^n)\equiv(2)^{-n\pmod{10}}+(2)^{3n-1\pmod{10}}-5\pmod{11}</math> | Hence, <math>a_n\equiv 6+(6^n)+6(8^n)\equiv(2)^{-n\pmod{10}}+(2)^{3n-1\pmod{10}}-5\pmod{11}</math> | ||
− | Therefore <math>a_{2018}\equiv4+8-5=7</math> | + | Therefore |
+ | |||
+ | <math>a_{2018}\equiv4+8-5=7</math> | ||
<math>a_{2020}\equiv1+6-5=2</math> | <math>a_{2020}\equiv1+6-5=2</math> |
Revision as of 11:36, 7 March 2020
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.