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Answer = 8.

The only small cubes with paint on three faces are those that occupied the corners of the original cube.

Q: How many small cubes will have paint on three faces if each cube is 3 inches and painted and cut into 27 smaller cubes that are 1 inch on each side?

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Only the 8 corner cubelets of the original large cube will have three painted faces.

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.....

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.

The colors don't matter. And I'm having to assume from your description that you mean that the smaller cubes have sides of length 2 cm long. Thus, there are 64 cubes. 8 have three sides painted. 24 have two sides painted. 24 have one side painted. 8 have no sides painted.

Related questions

0 sides are not painted! So all the cubes have three faces painted!

8

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).

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.

Only the 8 corner cubelets of the original large cube will have three painted faces.

Assuming all faces need to be painted, the question can be written as "how many faces on 43 cubes." There are 6 faces on each cube. Thus the answer can be got from the multiplication 43x6. Doing this gives us 258, so the number of faces that would need to be painted is 258.

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.....

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.

54

The colors don't matter. And I'm having to assume from your description that you mean that the smaller cubes have sides of length 2 cm long. Thus, there are 64 cubes. 8 have three sides painted. 24 have two sides painted. 24 have one side painted. 8 have no sides painted.

The colors don't matter. And I'm having to assume from your description that you mean that the smaller cubes have sides of length 2 cm long. Thus, there are 64 cubes. 8 have three sides painted. 24 have two sides painted. 24 have one side painted. 8 have no sides painted.