Posted by Jiyeon on Saturday, March 10, 2012 at 8:13pm.
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.
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
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