There would be 8 will 3 faces painted red, 24 with 2 faces painted red, 24 with 1 face painted red, and 8 that have no faces painted red.
When a cube is painted on all six faces and then sliced into 27 smaller cubes (which creates a 3x3x3 arrangement), the smaller cubes that are painted on only one face are those located in the center of each face of the larger cube. There are 6 faces on the cube, and each face has 1 center cube that is painted only on one side. Therefore, there are a total of 6 smaller cubes that are painted on only one face.
This is kind of neat. I drew a three dimensional cube. Then divided the faces into 16 squares per face. Then started counting which of those little cubes would have a painted face. I get: 1 red face - 24 cubes 2 red faces - 24 cubes 3 red faces - 8 cubes 4, 5, & 6 red faces - 0 cubes
Draw a three dimensional cube, showing three faces. Divide each face into 6 x 6 =36 squares, representing the faces of the little cubes. Then by inspection it is obvious that no little cube can have more then three painted faces and they will be at the eight corners. Only the edge cubes (not counting the corners) can have two painted faces and there are 48 of then. Finally, only the inner cubes on the surfaces will have one painted face and there are 96 of them. The other 64 cubes are in the inner core and will have no painted surfaces. SO: 6 = 0 ; 5 = 0; 4 =0 ; 3 = 8 ; 2 = 48 ; 1 = 96 . uuuuuu.....
I think the answer is 18 becausea cube has 6 face and 3 cubes have 18 faces.
It depends on how the rectangular solid is built. If the side of a cube is "a", then your original solid could be a x a x 54a, or a x 2a x 27a, or 2a x 3a x 9a or one of quite a few others, and the answer depends on the one you specify.
When a cube is painted on all six faces and then sliced into 27 smaller cubes (which creates a 3x3x3 arrangement), the smaller cubes that are painted on only one face are those located in the center of each face of the larger cube. There are 6 faces on the cube, and each face has 1 center cube that is painted only on one side. Therefore, there are a total of 6 smaller cubes that are painted on only one face.
I am assuming that your question states that the top and sides are painted red but the bottom isn't. If so, the answer is:- There are no cubes with 4 faces painted, the most that can be painted is 3 for the ones on the corners. There are 4 corners at the top of the cube that will have their top and 2 sides painted. Therefore there are 4 cubes with 3 painted. The cubes at the corners on the second and third row down will have 2 faces painted, as will the middle cubes on the top row so there are 12 of them in this puzzle. The cube in the middle of each painted face will have just one face painted so this is 5 (assuming the bottom isn't painted). The cubes in the middle of the bottom row will also have one face painted. This brings the total to 9 That is the total of cubes that have paint on them... 4+12+9 = 25 There are 27 cubes in your puzzle so only 2 have no painted faces. The cube right in the middle and the cube in the middle of the bottom layer.
54
All the little cubes except the one at the middle will have 1, 2 or 3 faces with paint on them, ie 26 of them will have some paint on them: 0 faces painted - 1 cube 1 face painted - 6 cubes (the centre ones of each side) 2 faces painted - 12 cubes (the ones in the centre of where two sides meet) 3 faces painted - 8 cubes (the ones in the corners).
This is kind of neat. I drew a three dimensional cube. Then divided the faces into 16 squares per face. Then started counting which of those little cubes would have a painted face. I get: 1 red face - 24 cubes 2 red faces - 24 cubes 3 red faces - 8 cubes 4, 5, & 6 red faces - 0 cubes
Draw a three dimensional cube, showing three faces. Divide each face into 6 x 6 =36 squares, representing the faces of the little cubes. Then by inspection it is obvious that no little cube can have more then three painted faces and they will be at the eight corners. Only the edge cubes (not counting the corners) can have two painted faces and there are 48 of then. Finally, only the inner cubes on the surfaces will have one painted face and there are 96 of them. The other 64 cubes are in the inner core and will have no painted surfaces. SO: 6 = 0 ; 5 = 0; 4 =0 ; 3 = 8 ; 2 = 48 ; 1 = 96 . uuuuuu.....
I think the answer is 18 becausea cube has 6 face and 3 cubes have 18 faces.
It depends on how the rectangular solid is built. If the side of a cube is "a", then your original solid could be a x a x 54a, or a x 2a x 27a, or 2a x 3a x 9a or one of quite a few others, and the answer depends on the one you specify.
8
Cubes have 6 faces (as well as 12 edges and 8 vertices).
Cube root of 64 is 4 therefore cube is 4 x 4 x 4. Each side of the cube has 16 visible small cube faces, 4 of which are corners (3 green sides) and 8 more are edges (2 green sides), so there are only four small cubes with one green face on each of the six sides of the large cube. Your answer is therefore 24.
Cubes are 3 dimensional figures and are equal on all edges. Therefore if one face is not a rectangle, then no face is a rectangle. Basically no.