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A hundred identical cubic boxes are currently arranged in

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A hundred identical cubic boxes are currently arranged in

by AAPL » Tue Feb 12, 2019 4:34 am

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Magoosh

A hundred identical cubic boxes are currently arranged in four cubes: a single cubic box, a 2 x 2 x 2 cube, a 3 x 3 x 3 cube, and a 4 x 4 x 4 cube. These four are not touching each other. All outward faces are painted and all inward faces are not painted. These four cubes are going to be dismantled and reassembled as a flat 10 x 10 square. The top and all the edges of this 10 x 10 square must be painted, but there is no requirement for paint on the bottom. How many individual faces will have to be painted to accommodate the requirements of this new design?

A. 0
B. 5
C. 9
D. 16
E. 27

OA C
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by Scott@TargetTestPrep » Wed Feb 13, 2019 6:33 pm
AAPL wrote:Magoosh

A hundred identical cubic boxes are currently arranged in four cubes: a single cubic box, a 2 x 2 x 2 cube, a 3 x 3 x 3 cube, and a 4 x 4 x 4 cube. These four are not touching each other. All outward faces are painted and all inward faces are not painted. These four cubes are going to be dismantled and reassembled as a flat 10 x 10 square. The top and all the edges of this 10 x 10 square must be painted, but there is no requirement for paint on the bottom. How many individual faces will have to be painted to accommodate the requirements of this new design?

A. 0
B. 5
C. 9
D. 16
E. 27

OA C

First let's determine the number of cubes that are already painted with at least one face:

The single cube, all 8 of the 2 x 2 x 2 cube, 26 of the 3 x 3 x 3 cube (since only 1 interior cube is not painted), and 56 of the 4 x 4 x 4 cube (since only 2^3 = 8 interior cubes are not painted).

Therefore, so far we have 1 + 8 + 26 + 56 = 91 cubes are painted with at least one face. So we need 9 more cubes to be painted so that all 100 cubes will be painted with at least one face.

Next, we need to make sure that we have enough cubes with two and three faces painted. Note that when the cubes are reassembled as a flat 10 x 10 square, the four corner cubes must have three faces painted. Also, of the remaining 10 - 2 = 8 cubes on each edge, 8 x 4 = 24 cubes must have two faces painted. The single cube together with three cubes from the 2 x 2 x 2 cube have at least three faces painted; thus these cubes can be used on the corners of the flat 10 x 10 square. We have 8 - 3 = 5 cubes with three faces painted remaining from the 2 x 2 x 2 cube. On the 3 x 3 x 3 cube, we have 26 cubes painted and only 6 of these cubes have only one side painted (the cubes that are located at the center of each face); therefore 26 - 6 = 20 cubes on the 3 x 3 x 3 cube have at least 2 sides painted. Together with the 5 cubes from the 2 x 2 x 2 cube, we have more than 24 cubes with at least 2 sides painted; therefore we only need the 9 faces that we determined earlier to be painted.

Answer: C

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by swerve » Thu Feb 14, 2019 11:28 am
Newly Assembled Cube - 100 Total Cubes (4 need 3 sides painted, 32 need 2 sides painted, 64 need 1 side painted)

1x1x1 cube has 1 cube with 6 sides painted
2x2x2 cube has 8 cubes with 3 sides painted
3x3x3 cube has 8 cubes with 3 sides painted, 12 cubes with 2 sides painted and 6 cubes with 1 side painted
4x4x4 cube has 8 cubes with 3 sides painted, 24 cubes with 2 sides painted and 24 cubes with 1 side painted

This leaves an excess of 9 cubes with 1 side painted required. C
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