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>. | ||
− | ==Solution== | + | ==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. | When given a sequence problem, one good thing to do is to check if the sequence repeats itself or if there is a pattern. | ||
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Our answer is <math>7</math> • <math>2</math> • <math>8</math> <math>= \boxed{112}</math>. | Our answer is <math>7</math> • <math>2</math> • <math>8</math> <math>= \boxed{112}</math>. | ||
+ | |||
+ | ==Solution 2 (Overkill) == | ||
+ | Notice that the characteristic polynomial of this is <math>x^3-4x^2-4x-4\equiv 0\pmod{11}</math> | ||
+ | |||
+ | Then since <math>x\equiv1</math> is a root, using Vieta's formula, the other two roots <math>r,s</math> satisfy <math>r+s\equiv3</math> and <math>rs\equiv4</math>. | ||
+ | |||
+ | Let <math>r=7+d</math> and <math>s=7-d</math>. | ||
+ | |||
+ | We have <math>49-d^2\equiv4</math> so <math>d\equiv1</math>. We found that the three roots of the characteristic polynomial are <math>1,6,8</math>. | ||
+ | |||
+ | Now we want to express <math>a_n</math> in an explicit form as <math>a(1^n)+b(6^n)+c(8^n)\pmod{11}</math>. | ||
+ | |||
+ | Plugging in <math>n=0,1,2</math> we get | ||
+ | <math>(*)</math><math>a+b+c\equiv2,</math> | ||
+ | |||
+ | <math>(**)</math><math>a+6b+8c\equiv5,</math> | ||
+ | |||
+ | <math>(***)</math><math>a+3b+9c\equiv8</math> | ||
+ | |||
+ | <math>\frac{(***)-(*)}{2}</math><math>\implies b+4c\equiv3</math> and <math>(***)-(**)</math><math>\implies -3b+c\equiv3</math> | ||
+ | |||
+ | so <math>a\equiv6,</math> <math>b\equiv1,</math> and <math>c\equiv6</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> | ||
+ | |||
+ | <math>a_{2020}\equiv1+6-5=2</math> | ||
+ | |||
+ | <math>a_{2022}\equiv3+10-5=8</math> | ||
+ | |||
+ | And the answer is <math>7\times2\times8=\boxed{112}</math> | ||
==See Also== | ==See Also== |
Revision as of 12:33, 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 |
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