## Cubes and Dices

The questions based on cubes, test the ability of the students to visualize and deal with the multidimensional problems/ situations.

#### Cube

A three dimensional figure having 6 square faces, 8 vertices and 12 edges with equal length, breadth and height is called cube.

**In a cube: **

Number of vertices V = 8

Number of Faces F = 6

Number of Edges E = 12

We know that

Euler’s formula = F + V = E + 2 |

#### Cuboid

A three dimensional figure in which length, breadth and height are different.

In cuboid 6 rectangular faces, 8 vertices and 12 edges.

Total number of small cubes = l × b × h |

#### Important Formula

Total number of small cube = n³ |

Where n = Number of rows/columns

1 side colored small cubes = 6 (n - 2)² |

2 sides colored small cubes = 12 (n - 2)² |

3 sides colored small cubes = 8 |

4 sides colored small cubes = 0 |

Colorless/Unseen cubes = (n - 2)³ |

**Ex-** A cube having dimension 5 cm has been transformed into smaller cube having dimension 1 cm. Then whole block is colored with pink color . Find

(1). Total number of small cubes

(2). 1 side colored small cubes

(3). 2 sides colored small cubes

(4). 3 sides colored small cubes

(5). 4 sides colored small cubes

(6). colorless cubes

**Solution:**- (1). Total number of small cubes = n^{3} = 5^{3} = 125

(2). 1 side colored small cubes = 6(n - 2)^{2}

= 6( 5 - 2 )^{2}

= 6 × 3^{2}

= 6 × 9

= 54

(3). 2 sides colored small cubes = 12(n - 2)

= 12 ( 5 - 2 ) = 12 × 3 = 36

(4). 3 sides colored small cubes = 8

(5). 4 sides colored small cubes = 0

(6). Colorless cubes = ( n - 2 )^{3}

= ( 5 - 2 )^{3} = 3^{3} = 27

**Direction:** 216 cubes of similar size are arranged in the form of the bigger cube (6 cubes on each side, i.e.., 6 x 6 x 6) all the exposed surfaces are painted.

**Ex-** How many of the cubes have 0 faces painted?

(1) 64

(2) 125

(3) 27

(4) None of these**Solution**:-Total number of cubes = n^{3}

or, 216 = n^{3}

∴ n = 6

Number of colorless cubes = ( n - 2 )^{3}

Number of the cubes with 0 faces painted is = (6 - 2)^{3} = 4^{3} = 64

• If we make a cut in one plane then we divide the cube in two parts (as shown in fig 1), if we make 2 cuts in same plane then we will get 2 + 1 = 3 pieces (as shown in fig 2) similarly if we make ‘p’ cuts in one plane then we will get ‘p + 1’ pieces as shown in Fig 3.

• If we make a cut in 2nd plane then we divide the cube in 2(p + 1) parts (as shown in fig 4), if we make 2 cuts in same plane then we will get (2 + 1) (p + 1) pieces (as shown in fig 5) similarly if we make ‘q’ cuts in 2nd plane then we will get ‘(p + 1) (q + 1) ’ pieces as shown in Fig 6.

• If we make a cut in 3rd plane then we divide the cube in 2(p + 1) (q + 1) parts (as shown in fig 7), if we make 2 cuts in same plane then we will get (2 + 1) (p + 1) (q + 1) pieces (as shown in fig 8) similarly if we make ‘q’ cuts in 3rd plane then we will get ‘(p + 1) (q + 1) (r + 1) ’ pieces as shown in Fig 9.

**Note: **‘a’ Number of cuts in one plane gives (a + 1) pieces, while ‘b’ cuts in another plane gives (b + 1) pieces, and ‘c’ cuts in 3rd plane gives (c + 1) pieces. Therefore a, b and c cuts in three dimensions will give (a + 1) (b + 1) (c + 1) pieces.

If number of cuts is given then maximum number of pieces can be obtain when a = b = c or these three are as close as possible while for minimum number of pieces can be obtain when all the cuts are made is in one plane.

**Example**- If total number of cuts is 10 then find the minimum and maximum number of pieces that can be obtained. **Solution:**- When all the cuts are in one plane then total number of pieces = 11 For maximum number of pieces a = 4, b = 3 and c = 3 then total number of pieces = 5 × 4 × 4 = 80

**Example**-If total number of pieces are 45 then find the possible number of cuts. **Solution:**- Since 45 = 1 × 1 × 45 = 1 × 3 × 15 = 1 × 9 × 5 = 3 × 3 × 5, and hence corresponding value of (a, b, c) = (0, 0, 44), (0, 2, 14), (0, 8, 4) & (2, 2, 4) and hence total number of cuts = 44, 16, 12 or 8

**Removal of a corner cube:** If a corner cube is removed and then all the exposed surface painted then changes due to the removal of a corner cube is as follows: (Lets take an example of 4 × 4 × 4 cube)

i. Total surface area will remain unchanged.

ii. Total number of cubes

iii. Number of cubes whose three face is painted will increase by 2 so number of such cubes is 10

iv. Number of cubes whose two face is painted will decrease by 3 so number of such cubes is 21

v. Number of cubes whose 1 face is painted will remain unchanged

vi. Number of cubes whose no face is painted will remain unchanged

**Example- **343 cubes of similar size are arranged in the form of a bigger cube (7 cubes on each side, i. e., 7 × 7 × 7) and kept at the corner of a room, all the exposed surfaces are painted then:

**1**.How many of the cubes have 0 faces painted?

(a) 64 (b) 125 (c) 240 (d) None of these

**2**. How many of the cubes have 2 faces painted?

(a) 14 (b) 18 (c) 16 (d) None of these

**3**. How many of the cubes have at most faces painted?

(a) 208 (b) 244 (c) 342 (d) None of these

**4**. How many of the cubes have at least 2 faces painted?

(a) 19 (b) 144 (c) 120 (d) None of these

**5**. How many of the cubes have 3 faces painted?

(a) 0 (b) 3 (c) 5 (d) None of these

**Solution:-** 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 side we will get 6 such cubes hence required number of cubes is 6 × 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 × 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.

**1**.(d From the above explanation number of the cubes with 0 faces painted is 216.

**2**. (b) From the above explanation number of the cubes with 2 faces painted is 18.

**3**. (c) From the above explanation number of the cubes with at most 2 faces painted is 216 + 108 + 18 = 342. Or else 343 -1 = 342

**4**. (a) From the above explanation number of the cubes with at least 2 faces painted is 18 + 1 = 19.

**5**. (d) From the above explanation number of the cubes with 3 faces painted is 1.