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

(Created page with "==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 remaind...")
 
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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>a_{n-2}</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==
 +
 
 +
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 <math>a_{0}</math>.
 +
 
 +
<math>a_{0} = 2</math>,
 +
<math>a_{1} = 5</math>,
 +
<math>a_{2} = 8</math>,
 +
<math>a_{3} = 5</math>,
 +
<math>a_{4} = 6</math>,
 +
<math>a_{5} = 10</math>,
 +
<math>a_{6} = 7</math>,
 +
<math>a_{7} = 4</math>,
 +
<math>a_{8} = 7</math>,
 +
<math>a_{9} = 6</math>,
 +
<math>a_{10} = 2</math>,
 +
<math>a_{11} = 5</math>,
 +
<math>a_{12} = 8</math>,
 +
<math>a_{13} = 5</math>
 +
 
 +
We can simplify the expression we need to solve to <math>a_{8}</math> • <math>a_{10}</math> • <math>a_{2}</math>.
 +
 
 +
Our answer is <math>7</math> • <math>2</math> • <math>8</math> <math>= \boxed{112}</math>.
 +
 
 +
==See Also==
  
 
{{AIME box|year=2018|n=II|num-b=1|num-a=3}}
 
{{AIME box|year=2018|n=II|num-b=1|num-a=3}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 08:30, 24 March 2018

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 $a_{2018}$ • $a_{2020}$ • $a_{2022}$.

Solution

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 $a_{8}$ • $a_{10}$ • $a_{2}$.

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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