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Difference between revisions of "2006 AIME A Problems/Problem 11"

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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 [[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.
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A collection of 8 [[cube (geometry) | cube]]s consists of one cube with [[edge]]-[[length]] <math> k </math> for each [[integer]] <math> k, 1 \le k \le 8. </math> A tower is to be built using all 8 cubes according to the rules:
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* Any cube may be the bottom cube in the tower.
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* The cube immediately on top of a cube with edge-length <math> k </math> must have edge-length at most <math> k+2. </math>  
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Let <math> T </math> be the number of different towers than can be constructed. What is the [[remainder]] when <math> T </math> is divided by 1000?
  
 
== Solution ==
 
== Solution ==

Revision as of 14:43, 25 September 2007

Problem

A collection of 8 cubes consists of one cube with edge-length $k$ for each integer $k, 1 \le k \le 8.$ A tower is to be built using all 8 cubes according to the rules:

  • Any cube may be the bottom cube in the tower.
  • The cube immediately on top of a cube with edge-length $k$ must have edge-length at most $k+2.$

Let $T$ be the number of different towers than can be constructed. What is the remainder when $T$ 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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