Difference between revisions of "2005 AIME I Problems/Problem 9"

 
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== Problem ==
 
== Problem ==
Twenty seven unit cubes are painted orange on a set of four faces so that two non-painted faces share an edge. The 27 cubes are randomly arranged to form a <math> 3\times 3 \times 3 </math> cube. Given the probability of the entire surface area of the larger cube is orange is <math> \frac{p^a}{q^br^c}, </math> where <math> p,q, </math> and <math> r </math> are distinct primes and <math> a,b, </math> and <math> c </math> are positive integers, find <math> a+b+c+p+q+r. </math>
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Twenty seven unit [[cube (geometry) | cube]]s are painted orange on a set of four [[face]]s so that two non-painted faces share an [[edge]]. The 27 cubes are randomly arranged to form a <math> 3\times 3 \times 3 </math> cube. Given the [[probability]] of the entire [[surface area]] of the larger cube is orange is <math> \frac{p^a}{q^br^c}, </math> where <math> p,q, </math> and <math> r </math> are distinct [[prime number | prime]]s and <math> a,b, </math> and <math> c </math> are [[positive integer]]s, find <math> a+b+c+p+q+r. </math>
  
 
== Solution ==
 
== Solution ==
 
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{{solution}}
 
== See also ==
 
== See also ==
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* [[2005 AIME I Problems/Problem 8 | Previous problem]]
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* [[2005 AIME I Problems/Problem 10 | Next problem]]
 
* [[2005 AIME I Problems]]
 
* [[2005 AIME I Problems]]

Revision as of 11:25, 30 October 2006

Problem

Twenty seven unit cubes are painted orange on a set of four faces so that two non-painted faces share an edge. The 27 cubes are randomly arranged to form a $3\times 3 \times 3$ cube. Given the probability of the entire surface area of the larger cube is orange is $\frac{p^a}{q^br^c},$ where $p,q,$ and $r$ are distinct primes and $a,b,$ and $c$ are positive integers, find $a+b+c+p+q+r.$

Solution

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See also