Nice problem for the math contests!
Assume the tunnel is through the middle square of one of the faces.
By symmetry, we only have to look at two rows of cubes, each parallel to the tunnel.
A. The corner row (count=4):
2 Exterior cubes with three faces painted
1 interior cube with two faces painted.
B. Interior row (count =4)
2 exterior cubes with 3 faces painted
1 interior cube with 2 faces painted.
Total:
f N
3 8
2 4
3 8
2 4
So there are 16 cubes with 3 faces painted and 8 cubes with 2 faces painted for a total of 24 cubes (=27-3).
Check: total number of faces painted = 16*3+8*2=64
From the original cube,
6*9+3*4-2(ends of tunnel)=64 checks.
Each cube below is made up of smaller cubes, but the large cubes are not solid. They had tunnels through them.
• The first cube originally had 27 small cubes, but the tunnel removed 3 cubes.
• The second cube originally had 64 small cubes, but two straight tunnels, 4 cubes deep, removed some cubes.
• The third cube, which originally had 125 small cubes, has 3 straight tunnels, five cubes deep from face to face.
The outside surfaces of these cube constructions have been painted including inside the tunnels and on the bottom. For each construction, how many small cubes have paint on 4 faces? 3 faces? 2 faces? 1 face? 0 face?
3 answers
I have the same exact question as this. But I don't understand your way of answering. Can you please help me with this?# of faces painted 3x3x3 4x4x4 5x5x5
4
3
2
1
0
4
3
2
1
0
There is no formulas involved, just counting will do.
To solve the problem by counting, we have to be organized.
Look at the view of the cube along the direction of the tunnel:
ACA
CTC
ACA
Each letter represents a cube of the 3x3 block.
The top row begins with type A (corner with three faces painted).
Between two corners, the type C has the top and bottom painted (it's above the tunnel) and the side facing us. So type C also has 3 sides painted.
On the second row, there are two type C's at each end, and the empty tunnel (T) in the middle.
The third row is identical to the top, just upside down.
If we count them, we have
4 type A (3 sides)
4 type C (3 sides)
The next (interior layer) has a map as follows:
BDB
DTD
BDB
The type B cubes at the corner have 2 faces painted, so have the middle ones (type D).
So we have
4 Type D (2 faces painted)
4 Type B (2 faces painted)
The far face is identical to the first, so again:
4 type A (3 sides)
4 type C (3 sides)
Summing it all up, we have
16 cubes with 3 faces painted, and
8 cubes with 2 faces painted as we had above.
See following sketch to help understand.
http://img10.imageshack.us/img10/3991/1331428434.jpg
To solve the problem by counting, we have to be organized.
Look at the view of the cube along the direction of the tunnel:
ACA
CTC
ACA
Each letter represents a cube of the 3x3 block.
The top row begins with type A (corner with three faces painted).
Between two corners, the type C has the top and bottom painted (it's above the tunnel) and the side facing us. So type C also has 3 sides painted.
On the second row, there are two type C's at each end, and the empty tunnel (T) in the middle.
The third row is identical to the top, just upside down.
If we count them, we have
4 type A (3 sides)
4 type C (3 sides)
The next (interior layer) has a map as follows:
BDB
DTD
BDB
The type B cubes at the corner have 2 faces painted, so have the middle ones (type D).
So we have
4 Type D (2 faces painted)
4 Type B (2 faces painted)
The far face is identical to the first, so again:
4 type A (3 sides)
4 type C (3 sides)
Summing it all up, we have
16 cubes with 3 faces painted, and
8 cubes with 2 faces painted as we had above.
See following sketch to help understand.
http://img10.imageshack.us/img10/3991/1331428434.jpg