Chapter 9.1, Problem 107E

a.

To determine

Complete the table.

Answer to Problem 107E

  $\begin{array}{ccccc}\begin{array}{l}\text{Number of }\\ \text{Blue Cube Faces}\end{array}& 0& 1& 2& 3\\ 3\times 3\times 3& 1& 6& 12& 8\end{array}$

Explanation of Solution

Given information:

A $3\times 3\times 3$ cube is made up of $27$ unit cubes (a unit cube has a length, width, and height of $1$ unit), and only the faces of each cube that are visible are painted blue, as shown in the figure.

Complete the table to determine how many unit cubes of the $3\times 3\times 3$ cube have 0 blue faces, $1$ blue face, $2$ blue faces, and $3$ blue faces

  $\begin{array}{ccccc}\begin{array}{l}\text{Number of }\\ \text{Blue Cube Faces}\end{array}& 0& 1& 2& 3\\ 3\times 3\times 3& & & & \end{array}$

Calculation:

The unit cube at centre does not have any face blue.

Hence, the numbers of unit cubes with no blue face is $1$ .

The middle unit cube on each face has only one blue face.

There are $6$ unit cubes each ahving only one blue face.

Hence, the numbers of unit cube with only one blue face are, $6$

There are $12$ unit cubes which have two blue faces. They are located at the middle of the edges of the face.

Rest of the ccubes has three blue faces.

Hence, total numbers of cubes having three blue faces are, $8$ .

Hence, the table is as follows,

  $\begin{array}{ccccc}\begin{array}{l}\text{Number of }\\ \text{Blue Cube Faces}\end{array}& 0& 1& 2& 3\\ 3\times 3\times 3& 1& 6& 12& 8\end{array}$

Hence, the table completed.

b.

To determine

Complete the table for the other cubes.

Answer to Problem 107E

  $\begin{array}{ccccc}\begin{array}{l}\text{Number of }\\ \text{Blue Cube Faces}\end{array}& 0& 1& 2& 3\\ 4\times 4\times 4& 8& 24& 24& 8\\ 5\times 5\times 5& 27& 54& 36& 8\\ 6\times 6\times 6& 64& 96& 48& 8\end{array}$

Explanation of Solution

Given information:

Repeat part (a) for a $4\times 4\times 4$ cube, a $5\times 5\times 5$ cube, and a $6\times 6\times 6$ cube.

Calculation:

The unit cube at centre does not have any face blue.

Hence, the numbers of unit cubes with no blue face is $1$ .

The middle unit cube on each face has only one blue face.

There are $6$ unit cubes each ahving only one blue face.

Hence, the numbers of unit cube with only one blue face are, $6$

There are $12$ unit cubes which have two blue faces. They are located at the middle of the edges of the face.

Rest of the ccubes has three blue faces.

Hence, total numbers of cubes having three blue faces are, $8$ .

For the other cubes the table is as follows,

  $\begin{array}{ccccc}\begin{array}{l}\text{Number of }\\ \text{Blue Cube Faces}\end{array}& 0& 1& 2& 3\\ 4\times 4\times 4& 8& 24& 24& 8\\ 5\times 5\times 5& 27& 54& 36& 8\\ 6\times 6\times 6& 64& 96& 48& 8\end{array}$

Hence, the table completed.

c.

To determine

Observe the pattern.

Answer to Problem 107E

  $\boxed{a_n={\left(n-2\right)}^3}$ , $\boxed{a_n=6{\left(n-2\right)}^2}$ and $\boxed{a_n=12\left(n-2\right)}$ .

Explanation of Solution

Given information:

What type of pattern do you observe?

Calculation:

Following pattern is observed,

Number of cubes with no blue faces:

  $1,8,27,64$

This is simplified as the following sequence,

  ${\left(3-2\right)}^3,{\left(4-2\right)}^3,{\left(5-2\right)}^3,{\left(6-2\right)}^3$

Therefore, the $n^{th}$ terms is,

  $a_n={\left(n-2\right)}^3$

Numbers of cube with one blue faces:

  $6,24,5,96$

This is simplified as the following sequence,

  $6.{\left(3-2\right)}^2,6.{\left(4-2\right)}^2,6.{\left(5-2\right)}^2,6.{\left(6-2\right)}^2$

Hence, the $n^{th}$ term is,

  $a_n=6{\left(n-2\right)}^2$

Numbers of cube with two blue faces:

  $12,24,36,48$

This is simplified as the following sequence,

  $12\left(3-2\right),12\left(4-2\right),12\left(5-2\right),12\left(6-2\right)$

Hence, the $n^{th}$ term is,

  $a_n=12\left(n-2\right)$

d.

To determine

Complete the table for an $n\times n\times n$ cube.

Answer to Problem 107E

  $\begin{array}{ccccc}\begin{array}{l}\text{Number of }\\ \text{Blue Cube Faces}\end{array}& 0& 1& 2& 3\\ n\times n\times n& {\left(n-2\right)}^3& 6{\left(n-2\right)}^2& 12\left(n-2\right)& 8\end{array}$

Explanation of Solution

Given information:

Write formulas you could use to repeat part (a) for an $n\times n\times n$ cube.

Calculation:

The unit cube at centre does not have any face blue.

Hence, the numbers of unit cubes with no blue face is $1$ .

The middle unit cube on each face has only one blue face.

There are $6$ unit cubes each ahving only one blue face.

Hence, the numbers of unit cube with only one blue face are, $6$

There are $12$ unit cubes which have two blue faces. They are located at the middle of the edges of the face.

Rest of the ccubes has three blue faces.

Hence, total numbers of cubes having three blue faces are, $8$ .

Writing all these in a table we get,

  $\begin{array}{ccccc}\begin{array}{l}\text{Number of }\\ \text{Blue Cube Faces}\end{array}& 0& 1& 2& 3\\ n\times n\times n& {\left(n-2\right)}^3& 6{\left(n-2\right)}^2& 12\left(n-2\right)& 8\end{array}$