How many of the cubes have at most faces painted?

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.

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.
Case (i): When red and blue are opposite to each other then from one face we will get 6 x 6 = 36 cubes bot out of them 2 x 6 cubes from common edge with green painted face is common so number of cubes are 2 x 6 x 6 - 2 x 6 = 60
Case (ii): When red and blue are adjacent to each other then green is either adjacent to these or opposite to any one of red or blue, in 1^st condition number of cubes is 2 x 6 x 6 - 2 x 6 - 11 = 55 cubes or in 2^nd condition 2 x 6 x 6 -6 - 6 = 60, required number of cubes is 55 or 60

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

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

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.
Total number of Cubes left after 7^thstep in (N - 7)³ in the form of (N - 7) x (N - 7) x (N - 7) cubes.
And out of these number of cubes whose two sides are painted is given by three edges with each edge has (N - 8) so total number of cubes is 3 x (N - 8)
From the given information 3(N - 8) = 21 or N = 15
Number of cubes removed in 3^rd step (i.e with colour 3) is = 3(N - 2)² - 3(N - 2) + 1 = 469

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.
The required number of cubes must be equal to difference between two positive integer
Since 64 - 27 = 37
125 - 27 = 98
125 - 64 = 61

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.
Number of cubes with only face is painted with colour 1 is 3(N - 2)(N - 1) = 3N² - 9N + 6
Number of cubes with only face is painted with colour 2 is 3 (N - 3)(N - 2) = 3N² - 15 N + 18
From the given condition (3N² - 9N + 6) + (3N² - 15 N + 18) = 6N² - 24N + 24 = 150 from this we will get N = 7.
Number of cubes left after step 3 is 4 x 4 x 4 = 64
When all the exposed faces are painted with colour 4 then number of cubes with only one face painted is 3 x 2 x 3 = 18
From the observation for 1^st step we have seen that number of cubes is 3(N - 2)(N - 1) or in other words 3 times the product of 2 consecutive integer that is satisfied only by 18 which is 3 times of 2 x 3.