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A cube is painted black before being sliced through equally 2 times in all directions. Find the number of cubes with none of their faces painted black, with 1 of their faces painted black, with 2 of their faces painted black and with 3 of their faces painted black.
Given :
A cube is painted and cut equally 2 times in all directions.
To find :
Number of cubes with none of their faces painted black,
Number of cubes with 1 of their faces painted black,
Number of cubes with 2 of their faces painted black and
Number of cubes with 3 of their faces painted black.
Solution :
- For a cube of side $n\times n\times n$ painted on all sides which is uniformly cut into smaller cubes of dimensions $1\times 1\times 1$,
- Number of cubes with 0 side painted= $(n-2)^3$
- Number of cubes with 1 sides painted =$6(n - 2)^2$
- Number of cubes with 2 sides painted= $12(n-2)$
- Number of cubes with 3 sides painted= 8(always)
The given cube is cut twice equally along all the directions. This implies, each side of the cube is cut into $(2+2+1)=5$ equal parts.
Therefore,
Number of smaller cubes with none of their faces painted black = $(5-2)^3$=$(3)^3=27.$
Number of smaller cubes with one of their faces painted black = $6(5-2)^2$=$6(3)^2=6\times9=54.$
Number of smaller cubes with two of their faces painted black = $12(5-2)$=$12(3)=36.$
Number of smaller cubes with three of their faces painted black = $8$.
Updated on: 10-Oct-2022
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