Which of the following can be the number of cubes removed from the original N ³ number of cubes.
(i) 37
(ii) 61
(iii) 98

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.
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

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.

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

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.

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.

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.
After step 1 number of cubes with exactly 2 face painted is 4(N - 1) + (N - 2) = 5N - 6
Similarly after 2^nd step number of cubes with exactly 2 face painted is 5(N - 2) - 6 = 5N - 11
And after 3^rd step number of cubes with exactly 2 face painted is 5(N - 2) - 6 = 5N - 16
So total number of such cubes is 15N - 33 out of the given options only option B satisfy the given condition.

Addition of line increases to the element.

The circle at the corner of the square moves clock wise, where as the cross inside the square moves anti clock wise.

In one step, the arrow and the small line segment turn to other side of the more line and in the next figure.