How many cubes are painted with red or blue but not green?

Here on each face 6 x 6 = 36 cubes that are painted with one colour.
Case (i): when these three colour are adjacent to each other then from one face we will get 6 x 6 = 36 cubes but out them 6 x 3 = 18 cubes from common edge is common so number of cubes are 3 x 6 x 6 - 6 x 3 = 90
Case (ii): When red and blue are opposite to each other (or any two of the given three) then required number of cubes is 3 x 6 x 6 - 2 x 6 = 96

Here on each face 6 x 6 = 36 cubes that are painted with one colour.
Case (i): When red and blue are adjacent to each other then from one face we will get 6 x 6 = 36 cubes but out of them 6 cubes from common edge is common so number of cubes are 2 x 6 x 6 - 6 = 66
Case (ii): When red and blue are opposite to each other then required number of cubes is 2 x 6 x 6 = 72

Number of cubes with two face painted from the top side (Which is a square of 3 x 3 = 9 cubes ) is 4.
Number of cubes with two face painted from the 2^nd from top side (Which has four edges and edge has 3 such cubes) is 4 x 3 = 12.
Number of such cubes from vertical edges is 4 x 1 = 4
Number of such cubes from bottom face is 4 x 1 = 4
Hence total such cubes is 4 + 12 + 4 + 4 = 24

Number of cubes with three coloured face on the top side = 4
Number of cubes with three coloured face on the 2^nd from top side = 4
Number of cubes with three coloured face on the bottom side = 12
Total number of such cubes = 12 + 8 = 20

Number of cubes with no face painted is
105 - 34 - 24 - 20 = 27
Or else all the 3 x 3 x 3 inner cubes will remain coloured.

Here on each face 6 x 6 = 36 cubes that are painted with one colour.
From solution of previous questions statements (ii) and (iii) are correct.

Here on each face 6 x 6 = 36 cubes that are painted with one colour.
None of the cubes can be painted in four faces.

Out of 6 faces of 5 faces are exposed and those were painted.
Number of vertices with three faces exposed (Painted) is 4
Number of vertices with 2 faces exposed (Painted) is 4
Number of vertices with 1 faces exposed (Painted) is 0
Number of vertices with 0 faces exposed (Painted) is 0
Number of sides with 2 sides exposed (Painted) is 8
Number of sides with 1 sides exposed (Painted) is 4
Number of sides with no sides exposed (Painted) is 0
From the above observation:
Number of cubes with 3 faces Painted is 4
Number of cubes with 2 faces Painted is given by sides which is exposed from two sides, out of 8 such edges 4 vertical edges will give us 6 cubes per edge and 4 edges from top surface will give us 5 such cubes from each edge and required number of cubes is 6 x 4 + 4 x 5 = 44.
Number of cubes with 1 face Painted is given by faces which is exposed from one sides four vertical faces will give us 6 x 5 = 30 cubes per face and top face will give us 5 x 5 = 25 and required number of cubes is 30 x 4 + 25 x 1 = 145
Number of cubes with 0 face Painted is given by difference between total number of cubes - number of cubes with at least 1 face painted = 343 - 4 - 44 - 145 = 150
In other words number of cubes with 0 painted is 6 x 5 x 5 = 150
From the above explanation number of the cubes with 0 faces painted is 150.
From the above explanation number of the cubes with at most 2 faces painted is 150 + 145 + 44 = 339.
Or else 343 - 4 = 339

Out of 6 faces of 5 faces are exposed and those were painted.
Number of vertices with three faces exposed (Painted) is 4
Number of vertices with 2 faces exposed (Painted) is 4
Number of vertices with 1 faces exposed (Painted) is 0
Number of vertices with 0 faces exposed (Painted) is 0
Number of sides with 2 sides exposed (Painted) is 8
Number of sides with 1 sides exposed (Painted) is 4
Number of sides with no sides exposed (Painted) is 0
From the above observation:
Number of cubes with 3 faces Painted is 4
Number of cubes with 2 faces Painted is given by sides which is exposed from two sides, out of 8 such edges 4 vertical edges will give us 6 cubes per edge and 4 edges from top surface will give us 5 such cubes from each edge and required number of cubes is 6 x 4 + 4 x 5 = 44.
Number of cubes with 1 face Painted is given by faces which is exposed from one sides four vertical faces will give us 6 x 5 = 30 cubes per face and top face will give us 5 x 5 = 25 and required number of cubes is 30 x 4 + 25 x 1 = 145
Number of cubes with 0 face Painted is given by difference between total number of cubes - number of cubes with at least 1 face painted = 343 - 4 - 44 - 145 = 150
In other words number of cubes with 0 painted is 6 x 5 x 5 = 150
From the above explanation number of the cubes with 0 faces painted is 150.
From the above explanation number of the cubes with 3 faces painted is 4.

Consider the 1^st step, initial number of cubes N³ after removal of 1^st set of coloured cubes number of cubes left out is (N - 1)³ hence number of cubes removed in 1^st step (i.e with colour 1) is
N³ - (N - 1)³ = 3N² - 3N + 1
Similarly number of cubes removed in 2^nd step (i.e with colour 2) is
Similarly number of cubes removed in 3^rd step is (i.e with colour 3) and so on.
= 3(N - 1)² - 3(N - 1) + 1
Number of cubes remaining after 1^st step is (N - 1)³
Number of cubes remaining after 2^nd step is (N - 2)³ and so on.
From the given condition
Number of cubes removed in 3^rd step (i.e with colour 3) is = 3(N - 2)² - 3(N - 2) + 1 = 217 hence N = 11 So number of cubes with colour 5 is = 3(N - 4)² - 3(N - 4) + 1 = 127