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Difference between revisions of "2018 AIME II Problems/Problem 2"

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==Problem==
 
==Problem==
  
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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>.
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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} \cdot a_{2020} \cdot a_{2022}</math>.
  
 
==Solution 1==
 
==Solution 1==
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<math>a_{13} = 5</math>
 
<math>a_{13} = 5</math>
  
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We can simplify the expression we need to solve to <math>a_{8}</math> • <math>a_{10}</math> • <math>a_{2}</math>.
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We can simplify the expression we need to solve to <math>a_{8}\cdot a_{10} \cdot a_{2}</math>.
  
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Our answer is <math>7</math> • <math>2</math> • <math>8</math> <math>= \boxed{112}</math>.
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Our answer is <math>7 \cdot 2 \cdot 8</math> <math>= \boxed{112}</math>.
  
 
==Solution 2 (Overkill) ==
 
==Solution 2 (Overkill) ==
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Plugging in <math>n=0,1,2</math> we get
 
Plugging in <math>n=0,1,2</math> we get
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<math>(*)</math><math>a+b+c\equiv2,</math>
 
<math>(*)</math><math>a+b+c\equiv2,</math>
  
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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>
  
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Therefore <math>a_{2018}\equiv4+8-5=7</math>
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Therefore
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<math>a_{2018}\equiv4+8-5=7</math>
  
 
<math>a_{2020}\equiv1+6-5=2</math>
 
<math>a_{2020}\equiv1+6-5=2</math>

Latest revision as of 16:03, 12 January 2022

Problem

Let $a_{0} = 2$, $a_{1} = 5$, and $a_{2} = 8$, and for $n > 2$ define $a_{n}$ recursively to be the remainder when $4$($a_{n-1}$ $+$ $a_{n-2}$ $+$ $a_{n-3}$) is divided by $11$. Find $a_{2018} \cdot a_{2020} \cdot a_{2022}$.

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 $a_{0}$.

$a_{0} = 2$, $a_{1} = 5$, $a_{2} = 8$, $a_{3} = 5$, $a_{4} = 6$, $a_{5} = 10$, $a_{6} = 7$, $a_{7} = 4$, $a_{8} = 7$, $a_{9} = 6$, $a_{10} = 2$, $a_{11} = 5$, $a_{12} = 8$, $a_{13} = 5$

We can simplify the expression we need to solve to $a_{8}\cdot a_{10} \cdot a_{2}$.

Our answer is $7 \cdot 2 \cdot 8$ $= \boxed{112}$.

Solution 2 (Overkill)

Notice that the characteristic polynomial of this is $x^3-4x^2-4x-4\equiv 0\pmod{11}$

Then since $x\equiv1$ is a root, using Vieta's formula, the other two roots $r,s$ satisfy $r+s\equiv3$ and $rs\equiv4$.

Let $r=7+d$ and $s=7-d$.

We have $49-d^2\equiv4$ so $d\equiv1$. We found that the three roots of the characteristic polynomial are $1,6,8$.

Now we want to express $a_n$ in an explicit form as $a(1^n)+b(6^n)+c(8^n)\pmod{11}$.

Plugging in $n=0,1,2$ we get

$(*)$$a+b+c\equiv2,$

$(**)$$a+6b+8c\equiv5,$

$(***)$$a+3b+9c\equiv8$

$\frac{(***)-(*)}{2}$$\implies b+4c\equiv3$ and $(***)-(**)$$\implies -3b+c\equiv3$

so $a\equiv6,$ $b\equiv1,$ and $c\equiv6$

Hence, $a_n\equiv 6+(6^n)+6(8^n)\equiv(2)^{-n\pmod{10}}+(2)^{3n-1\pmod{10}}-5\pmod{11}$

Therefore

$a_{2018}\equiv4+8-5=7$

$a_{2020}\equiv1+6-5=2$

$a_{2022}\equiv3+10-5=8$

And the answer is $7\times2\times8=\boxed{112}$

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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