**Build your concepts to master the Problems on coloured Cubes**

Questions on coloured cubes are very common in any aptitude exam.

The question usually talks about a cube whose faces are painted and and then the cube is cut into smaller cubes. You are expected to tell the number of smaller cubes having zero, one, two or three faces painted.(More difficult variations of the question contains cubes with faces painted in different colours or some faces not painted at all)

Quite confusing and time taking if you don’t practice these questions well and more importantly don’t know the concept behind the solutions. We hope this post helps you visualise the concept involved in solving the questions.

**Let’s start with the basics..**

A Cube has 6 faces, 12 edges and 8 corners.

Now, let’s get into the subject matter.

**A cube painted on all faces is cut into n^3 smaller cubes.**

**1. The cubes in the corners of the bigger cube are painted on 3 faces.**(Pink cubes in the image above)

These are always 8 as they have to be at the corners for sure and there are only 8 corners in a cube irrespective of the dimensions or the number of smaller cubes it is cut into. So there’s no formula to derive here, just remember the number **8**.

**2. The cubes along the 12 edges except those at the corners are painted on 2 faces.**(Blue cubes in the image above)

Count all the cubes falling on edges but not at the corner.

For an ‘nxn’ cube the edge contains n cubes. We need to exclude the cubes at the corners coz they will be painted on three sides. So we have ‘n-2’ cubes on every edge painted on two faces. Further, there are 12 edges, hence totally 12*(n-2) cubes will be painted on two faces.

**For example:**

Let’s take a 3×3 cube. Every edge contains 3 cubes. But excluding the two cubes lying at the corners we are left with 3-2 = 1 cube which is painted on two faces. As there are 12 edges, the total no of cubes = 12*1 = 12

Using the formula (12*(n-2)) for the other cases:

4×4 cube: 12*(4-2) = 24

5×5 cube: 12*(5-2) = 36

**3. The cubes at the center of each face of the bigger cube are painted on one face.**(Green cubes in the image above)

For this case you have to count all the cubes on each face excluding the cubes lying on the edges.

**Example:**

Let’s work it out for a sa(i)mple case:

For an nxn parent cube there are n^2 smaller cubes on each face of the parent cube.Look at the diagram above, this is like an nxn matrix with n rows and n columns. Excluding the 4 deges the matrix becomes (n-2)x(n-2). Hence, there are (n-2)^2 cubes on this face which are painted on a single face. There 6 such faces in the bigger cube therefore there are totally 6*(n-2)^2 cubes with single face painted.

**Formulae:**

1. Number of small cubes **painted on 3 faces = 8** (You blindly tick this option in your exam without any calculation or thinking for a cube of any dimension 3×3, 4×4 etc)

2. Number of small cubes **painted only on 2 faces = 12(n-2)**

3. Number of small cubes **painted only on 1 face = 6(n-2)^2**

4. Number of small cubes **unpainted on all faces = (n-2)^3**

**Now, based on the above formulae you can further arrive at the following conclusions:** (These are some other frequently asked questions based on the same concept)

1. Number of small cubes with **atleast one face painted = 6(n-2)^2 + 12(n-2) + 8 **[No of cubes painted on one face + painted two faces + painted on three faces]

2. Number of small cubes with **atleast two faces painted = 12(n-2) + 8 **[No of cubes painted on two faces + painted on three faces]

We, will come up with some questions with detailed solutions on the above topic in our next post.

Your comments and feedback are most welcome.

**Happy Reading!!**

**Team CareerShapers**