Find the minimum number of cuts required to get 50 pieces.

For maximum number of cuts it has to be in one cut only, so number of cuts is 49

If 45 = 1 x 1 x 45 then we require only 44 cuts in one plane.
If 1 x 3 x 15 then we require 2 cuts in one plane and 14 cuts in other plane so total number of cuts is 2 + 14 = 16.
If 1 x 5 x 9 the we require 4 cuts in one plane and 8 cuts in other plane so total number of cuts is 4 + 8 = 12
If 3 x 3 x 5 then we require 2 cuts in one plane, 2 cuts in 2^nd plane and 4 cuts in 3^rd plane so total number of cuts is 2 + 2 + 4 = 8.

For maximum number of pieces cuts has to be 6, 7 and 7 and maximum number of pieces is (6 + 1)(7 + 1)(7 + 1) = 7 x 8 x 8 = 448.
Minimum number of pieces is 20 + 1 = 21.
Hence required ratio is 448:21

If total number of cut is 10 then for maximum number of pieces these cuts have to be well distributed in three planes. For 10 cuts, 3,3 and 4 is the distribution of cuts.
Hence total number of pieces is
(3 + 3)(3 + 1)(4 + 1) = 4 x 4 5 = 80

If total number of cuts is 10 then minimum number of pieces is 11 when cut is made in one plane only.

Since total number of cubes is hence in the formula we will substitute n = 6
Number of the cubes with 0 faces painted is (6 - 2)³ = 4³ = 64

Since total number of cubes is hence in the formula we will substitute n = 6
Number of the cubes with 2 faces painted is 6(6 - 2)²
= 6 x 16 = 96

Since total number of cubes is hence in the formula we will substitute n = 6
At most 2 faces painted means number of cubes with 0 face painted + number of cubes with 1 face painted + number of cubes with 2 face painted = 64 + 48 + 96 = 208

Since total number of cubes is hence in the formula we will substitute n = 6
At least 2 faces painted means number of cubes with 2 face painted + number of cubes with 3 face painted = 96 + 8 = 104.

Out of 6 faces of 3 faces are exposed and those were painted.
Number of vertices with three faces exposed (Painted) is 1
Number of vertices with 2 faces exposed (Painted) is 3
Number of vertices with 1 faces exposed (Painted) is 3
Number of vertices with 0 faces exposed (Painted) is 1
Number of sides with 2 sides exposed (Painted) is 3
Number of sides with 1 sides exposed (Painted) is 6
Number of sides with no sides exposed (Painted) is 3
From the above observation
Number of cubes with 3 faces Painted is 1
Number of cubes with 2 faces Painted is given by sides which is exposed from two sides and there are 3 such sides and from one we will get 6 such cubes hence required number of cubes is 6 x 3 = 18
Number of cubes with 1 face Painted is given by faces which is exposed from one sides and there are 3 such faces hence required number of cubes is 36 x 3 = 108
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 - 1- 18 - 108 = 216
In other words number of cubes with 0 painted is (7 - 1)³ = 216.
From the above explanation number of the cubes with 0 faces painted is 216.