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

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== Problem ==
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#REDIRECT [[2006 AIME I Problems/Problem 11]]
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:
 
 
 
* Any cube may be the bottom cube in the tower.
 
* The cube immediately on top of a cube with edge-length <math> k </math> must have edge-length at most <math> k+2. </math>
 
 
 
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 ==
 
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 ==
 
*[[2006 AIME II Problems]]
 
 
 
{{AIME box|year=2006|n=II|num-b=10|num-a=12}}
 
 
 
[[Category:Intermediate Algebra Problems]]
 

Latest revision as of 14:15, 23 August 2009

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