All faces of a 4-inch cube have been painted. If the cube is cut into 1-inch smaller cubes, how many of them have no paint on all their faces?
(With solution pls...) tnx
To solve this problem, we need to find the number of 1-inch smaller cubes that have no paint on any of their faces.
First, let's consider the original 4-inch cube. It has six faces, and each face has its own unique color.
When we cut the 4-inch cube into smaller 1-inch cubes, each face of the larger cube becomes a face of the smaller cubes. Therefore, all the smaller cubes that are created from a particular face will have the same color on that face.
Since all the faces of the original 4-inch cube are painted, we can conclude that each face of the smaller 1-inch cubes will be painted as well.
We are looking for the number of cubes that have no paint on any of their faces. This means that we need to count the smaller cubes that are completely surrounded by other cubes on all six sides.
If we imagine the 4-inch cube as consisting of a 2x2x2 arrangement of smaller cubes, the small cubes positioned along the edges have three faces exposed, and the small cubes positioned at the corners have two faces exposed. Therefore, they cannot have all of their faces unpainted.
The only smaller cubes that can have all of their faces unpainted are those located inside the larger cube. Taking into consideration that we have a 2x2x2 arrangement, there are 2 x 2 x 2 = 8 smaller cubes inside the 4-inch cube.
Thus, there are 8 smaller cubes that have no paint on any of their faces.