Chapter 9.1, Problem 107E

(a)

To determine

The unit cubes of $3\times 3\times 3$ cube have 0 blue faces, 1 blue face, 2 blue faces and 3 blue faces.

Answer to Problem 107E

The table is as follows,

  

Explanation of Solution

Given information:

The faces of each cube that are visible are painted blue as shown below,

  

Formula used:

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

Calculation:

From the following figure we have,

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

Therefore, 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 having only one blue faces.

Therefore, 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 cubes has three blue faces.

Therefore, total numbers or cubes having three blue faces are, 8.

Therefore, the table is as follows,

  

Conclusion:

The table is as follows,

  

(b)

To determine

The unit cubes for $4\times 4\times 4$ cube, $5\times 5\times 5$ cube and $6\times 6\times 6$ cube.

Answer to Problem 107E

The table is as follows,

  

Explanation of Solution

Given information:

The faces of each cube that are visible are painted blue as shown below,

  

Formula used:

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

Calculation:

For the other cubes the table is as follows,

  

Conclusion:

The table is as follows,

  

(c)

To determine

The formulas for an $n\times n\times n$ cube.

Answer to Problem 107E

The n^th term is,

  $a_n=12\left(n-2\right)$

Explanation of Solution

Given information:

The faces of each cube that are visible are painted blue as shown below,

  

Formula used:

The n^th term is,

  $a_n={\left(n-2\right)}^3$

Calculation:

Following pattern is observed,

Numbers of cubes with no blue face:

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 term is,

  $a_n={\left(n-2\right)}^3$

Numbers of cube with one blue face:

6,24,54,96

This is simplified as the following sequence,

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

Therefore, 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,

Therefore, the n^th term is,

  $a_n=12\left(n-2\right)$

  

Conclusion:

The n^th term is,

  $a_n=12\left(n-2\right)$