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
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. 
