What is the least possible number of piece which have at most one colour on them?

For least number of cuts 120 = 4 x 5 x 6 i.e number of cuts must be 3, 4 and 5 in three planes in this case number of cubes on a face is either 6 x 5 = 30 or 6 x 4 = 24 or 4 x 5 = 20 cubes . And number of cuboids on an edge is 4 or 5 or 6
Number of cuboids with no face painted is (4 - 2)(5 - 2)(6 - 2) = 2 x 3 x 4 = 24

Let us see the changes due to removal of cube from corner-
Number of vertices with three faces exposed (Painted) is 7 + 3 = 10
Number of Cubes with 2 sides exposed (Painted): In general one edge gives us 4 (n - 2 in general case) cubes with two face painted but in this case out of 12 edges only 9 edges will give us 4 cubes in one edge and remaining 3 edges will give us 3 cubes from one edge, hence total number of edge is 9 x 4 + 3 x 3 = 45
Number of Cubes with 1 side exposed (Painted): It will remain same as normal case i.e. 6(4²) = 96
Number of Cubes with no sides exposed (Painted) is 4³ = 64
From the above observation:
No cubes are with 4 face painted.

Let us see the changes due to removal of cube from corner-
Number of vertices with three faces exposed (Painted) is 7 + 3 = 10
Number of Cubes with 2 sides exposed (Painted): In general one edge gives us 4 (n - 2 in general case) cubes with two face painted but in this case out of 12 edges only 9 edges will give us 4 cubes in one edge and remaining 3 edges will give us 3 cubes from one edge, hence total number of edge is 9 x 4 + 3 x 3 = 45
Number of Cubes with 1 side exposed (Painted): It will remain same as normal case i.e. 6(4²) = 96
Number of Cubes with no sides exposed (Painted) is 4³ = 64
From the above observation:
From the above explanation number of the cubes with at least 2 faces painted is 45 + 10 = 55.

Let us see the changes due to removal of cube from corner-
Number of vertices with three faces exposed (Painted) is 7 + 3 = 10
Number of Cubes with 2 sides exposed (Painted): In general one edge gives us 4 (n - 2 in general case) cubes with two face painted but in this case out of 12 edges only 9 edges will give us 4 cubes in one edge and remaining 3 edges will give us 3 cubes from one edge, hence total number of edge is 9 x 4 + 3 x 3 = 45
Number of Cubes with 1 side exposed (Painted): It will remain same as normal case i.e. 6(4²) = 96
Number of Cubes with no sides exposed (Painted) is 4³ = 64
From the above observation:
From the above explanation number of the cubes with at most 2 faces painted is 64 + 96 + 45 = 205.
Or else 215 - 10 = 205

Let us see the changes due to removal of cube from corner-
Number of vertices with three faces exposed (Painted) is 7 + 3 = 10
Number of Cubes with 2 sides exposed (Painted): In general one edge gives us 4 (n - 2 in general case) cubes with two face painted but in this case out of 12 edges only 9 edges will give us 4 cubes in one edge and remaining 3 edges will give us 3 cubes from one edge, hence total number of edge is 9 x 4 + 3 x 3 = 45
Number of Cubes with 1 side exposed (Painted): It will remain same as normal case i.e. 6(4²) = 96
Number of Cubes with no sides exposed (Painted) is 4³ = 64
From the above observation:
From the above explanation number of the cubes with 2 faces painted is 45.

For least number of cuts 120 = 4 x 5 x 6 i.e number of cuts must be 3, 4 and 5 in three planes in this case number of cubes on a face is either 6 x 5 = 30 or 6 x 4 = 24 or 4 x 5 = 20 cubes . And number of cuboids on an edge is 4 or 5 or 6
In this case when k is maximum, one particular colour is used on there faces such that any two faces are adjacent to each other. Required number of cuboids will come from edges but not from vertex = 3 + 4 + 5 + 1 = 13

For least number of cuts 120 = 4 x 5 x 6 i.e number of cuts must be 3, 4 and 5 in three planes in this case number of cubes on a face is either 6 x 5 = 30 or 6 x 4 = 24 or 4 x 5 = 20 cubes . And number of cuboids on an edge is 4 or 5 or 6
Maximum number of cuboid with red colour is possible when cube is painted with red colour in 3 sides with minimum number of common edges (which is equal to 2)
Hence required maximum value is 6 (5 + 5 + 4 - 2) = 72
For minimum number of such cuboid Red colour is used only once and minimum number of cubes in that case is 20
Hence required ratio is 72 : 20 = 18 : 5

For least number of cuts 120 = 4 x 5 x 6 i.e number of cuts must be 3, 4 and 5 in three planes in this case number of cubes on a face is either 6 x 5 = 30 or 6 x 4 = 24 or 4 x 5 = 20 cubes . And number of cuboids on an edge is 4 or 5 or 6
In this case we have to use red and green twice and same colour should be on opposite faces then required cube is given by 4 edges (but not corner), maximum number of cubes one edge is 6 - 2 = 4 so required number of cubes is 4 x 4 = 16

Total no. of cubes = 5³ = 125,
Some cubes from different corners are removed and the number removed cubes are 2, 3, 4 and 4.
Remaining number of small cubes:
= 125 - 2 - 3 - 4 - 4 = 125 - 13 = 112

In any plane, leave 4 sides cube and select (3 x 3 x 3) inter section. But
the cubes 2 x 2 x 1 give 2 less cube because that part we are already removed.
No. of cubes = (3 x 3 x 3) - 2 = 25.