2005 Alabama ARML TST Problems/Problem 2: Difference between revisions
Thanks, Tempral! :D |
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==Solution== | ==Solution== | ||
The cube is | The cube is <math>4\times 4\times4</math>. Only the inside sub-cubes are unpainted, thus all but <math>8</math> are unpainted, leaving <math>64-8=56 </math>painted. | ||
==See also== | ==See also== | ||
{{ARML box|year=2005|state=Alabama|num-b=1|num-a=3}} | {{ARML box|year=2005|state=Alabama|num-b=1|num-a=3}} | ||
[[Category:Intermediate Combinatorics Problems]] | |||
Revision as of 11:42, 11 December 2007
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 
. Only the inside sub-cubes are unpainted, thus all but 
are unpainted, leaving 
painted.
See also
| 2005 Alabama ARML TST (Problems) | ||
| Preceded by: Problem 1 |
Followed by: Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
