Difference between revisions of "2005 Alabama ARML TST Problems/Problem 2"

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==Solution==
 
==Solution==
The cube is 4x4x4. Only the inside subcubes are unpainted, thus all but 8 are unpainted, leaving 64-8=56 painted.
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The cube is <math>4\times 4\times4</math>. Only the inside sub-cubes are unpainted, thus <math>8</math> are unpainted, leaving <math>64-8=56</math> painted.
  
 
==See also==
 
==See also==
*[[2005 Alabama ARML TST]]
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{{ARML box|year=2005|state=Alabama|num-b=1|num-a=3}}
*[[2005 Alabama ARML TST Problems/Problem 1 | Previous Problem]]
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*[[2005 Alabama ARML TST Problems/Problem 3 | Next Problem]]
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[[Category:Intermediate Combinatorics Problems]]

Latest revision as of 12:06, 26 January 2009

Problem

A large cube is painted green and then chopped up into 64 smaller congruent cubes. How many of the smaller cubes have at least one face painted green?

Solution

The cube is $4\times 4\times4$. Only the inside sub-cubes are unpainted, thus $8$ are unpainted, leaving $64-8=56$ painted.

See also

2005 Alabama ARML TST (Problems)
Preceded by:
Problem 1
Followed by:
Problem 3
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