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Difference between revisions of "2006 AIME A Problems/Problem 11"
(add solution, Template:AIME box, wikify, etc)
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== Problem ==
 
== Problem ==
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A sequence is defined as follows <math> a_1=a_2=a_3=1, </math> and, for all positive integers <math> n, a_{n+3}=a_{n+2}+a_{n+1}+a_n. </math> Given that <math> a_{28}=6090307, a_{29}=11201821, </math> and <math> a_{30}=20603361, </math> find the remainder when <math> \displaystyle \sum^{28}_{k=1} a_k </math> is divided by 1000.
+
A [[sequence]] is defined as follows <math> a_1=a_2=a_3=1, </math> and, for all positive [[integer]]s <math> n, a_{n+3}=a_{n+2}+a_{n+1}+a_n. </math> Given that <math> a_{28}=6090307, a_{29}=11201821, </math> and <math> a_{30}=20603361, </math> find the [[remainder]] when <math> \displaystyle \sum^{28}_{k=1} a_k </math> is divided by 1000.
  
 
== Solution ==
 
== Solution ==
 +
Define the sum as <math>x</math>. Notice that <math>a_n\ = a_{n + 3} - a_{n + 2} - a_{n + 1} </math>, so the sum will be:
 +
:<math>x = (a_4 - a_3 - a_2) + (a_5 - a_4 - a_3) + \ldots (a_{30} - a_{29} - a_{28}) + a_{28}</math>
 +
:<math>x = (a_4+ a_5 \ldots a_{30}) - (a_3 + a_4 + \ldots a_{29}) - (a_2 + a_3 + \ldots a_{28}) + a_{28} + (a_1 - a_1)</math>
 +
 +
The first two groupings almost completely cancel. The third resembles <math>x</math>.
 +
 +
:<math>x\ = a_1 - a_3 + a_{28} + a_{30} - x</math>
 +
:<math>2x\ = a_{28} + a_{30}</math>
 +
:<math>x\ = \frac{a_{28} + a_{30}}{2}</math>
 +
 +
<math>a_{28}</math> and <math>a_{30}</math> are both given; the last four digits of the sum is <math>3668</math>, and half of that is <math>1834</math>. Therefore, the answer is <math>834</math>.
  
 
== See also ==
 
== See also ==
 
*[[2006 AIME II Problems]]
 
*[[2006 AIME II Problems]]
 +
 +
{{AIME box|year=2006|n=II|num-b=10|num-a=12}}
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]

Revision as of 20:27, 6 February 2007

Problem

A sequence is defined as follows $a_1=a_2=a_3=1,$ and, for all positive integers $n, a_{n+3}=a_{n+2}+a_{n+1}+a_n.$ Given that $a_{28}=6090307, a_{29}=11201821,$ and $a_{30}=20603361,$ find the remainder when $\displaystyle \sum^{28}_{k=1} a_k$ is divided by 1000.

Solution

Define the sum as $x$. Notice that $a_n\ = a_{n + 3} - a_{n + 2} - a_{n + 1}$, so the sum will be:

$x = (a_4 - a_3 - a_2) + (a_5 - a_4 - a_3) + \ldots (a_{30} - a_{29} - a_{28}) + a_{28}$
$x = (a_4+ a_5 \ldots a_{30}) - (a_3 + a_4 + \ldots a_{29}) - (a_2 + a_3 + \ldots a_{28}) + a_{28} + (a_1 - a_1)$

The first two groupings almost completely cancel. The third resembles $x$.

$x\ = a_1 - a_3 + a_{28} + a_{30} - x$
$2x\ = a_{28} + a_{30}$
$x\ = \frac{a_{28} + a_{30}}{2}$

$a_{28}$ and $a_{30}$ are both given; the last four digits of the sum is $3668$, and half of that is $1834$. Therefore, the answer is $834$.

See also

2006 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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