EXERCISE 6.1 SOLUTIONS

1. Which of the following numbers are not perfect cubes?

(i) 216 (ii) 128 (iii) 1000 (iv) 100 (v) 46656

(i) 216

Step 1: Prime factorization of 216

Divisor Number
2 216
2 108
2 54
3 27
3 9
3 3
1

Step 2: Prime factorization expression

\[216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 = 2^3 \times 3^3\]

Step 3: Analyze the prime factors

We can see that:

  • Prime factor 2 appears 3 times (a multiple of 3)
  • Prime factor 3 appears 3 times (a multiple of 3)

∴ 216 is a perfect cube.

(ii) 128

Step 1: Prime factorization of 128

Divisor Number
2 128
2 64
2 32
2 16
2 8
2 4
2 2
1

Step 2: Prime factorization expression

\[128 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^7\]

Step 3: Analyze the prime factors

We can see that:

  • Prime factor 2 appears 7 times
  • 7 is not a multiple of 3

∴ 128 is not a perfect cube.

(iii) 1000

Step 1: Prime factorization of 1000

Divisor Number
2 1000
2 500
2 250
5 125
5 25
5 5
1

Step 2: Prime factorization expression

\[1000 = 2 \times 2 \times 2 \times 5 \times 5 \times 5 = 2^3 \times 5^3\]

Step 3: Analyze the prime factors

We can see that:

  • Prime factor 2 appears 3 times (a multiple of 3)
  • Prime factor 5 appears 3 times (a multiple of 3)

∴ 1000 is a perfect cube.

(iv) 100

Step 1: Prime factorization of 100

Divisor Number
2 100
2 50
5 25
5 5
1

Step 2: Prime factorization expression

\[100 = 2 \times 2 \times 5 \times 5 = 2^2 \times 5^2\]

Step 3: Analyze the prime factors

We can see that:

  • Prime factor 2 appears 2 times (not a multiple of 3)
  • Prime factor 5 appears 2 times (not a multiple of 3)

∴ 100 is not a perfect cube.

(v) 46656

Step 1: Prime factorization of 46656

Divisor Number
2 46656
2 23328
2 11664
2 5832
2 2916
2 1458
3 729
3 243
3 81
3 27
3 9
3 3
1

Step 2: Prime factorization expression

\[46656 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 2^6 \times 3^6\]

Step 3: Analyze the prime factors

We can see that:

  • Prime factor 2 appears 6 times (a multiple of 3)
  • Prime factor 3 appears 6 times (a multiple of 3)

∴ 46656 is a perfect cube.

2. Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.

(i) 243 (ii) 256 (iii) 72 (iv) 675 (v) 100

(i) 243

Step 1: Prime factorization of 243

Divisor Number
3 243
3 81
3 27
3 9
3 3
1

Step 2: Prime factorization expression

\[243 = 3 \times 3 \times 3 \times 3 \times 3 = 3^3 \times 3^2 \]

Step 3:

3 is without triplet

To make it a perfect cube, We need to multiply it by 3 .

∴ The smallest number is 3.

(ii) 256

Step 1: Prime factorization of 256

Divisor Number
2 256
2 128
2 64
2 32
2 16
2 8
2 4
2 2
1

Step 2: Prime factorization expression

\[256 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^3 \times 2^3 \times 2^2 \]

Step 3:

2 is without triplet.

To make it a perfect cube, We need to multiply it by 2.

∴ The smallest number is 2.

(iii) 72

Step 1: Prime factorization of 72

Divisor Number
2 72
2 36
2 18
3 9
3 3
1

Step 2: Prime factorization expression

\[72 = 2 \times 2 \times 2 \times 3 \times 3 = 2^3 \times 3^2\]

Step 3:

3 is without triplet

To make it a perfect cube, We need to multiply it by 3.

∴ The smallest number is 3.

(iv) 675

Step 1: Prime factorization of 675

Divisor Number
3 675
3 225
3 75
5 25
5 5
1

Step 2: Prime factorization expression

\[675 = 3 \times 3 \times 3 \times 5 \times 5 = 3^3 \times 5^2\]

Step 3:

5 is without triplet

To make it a perfect cube, We need to multiply it by 5.

∴ The smallest number is 5.

(v) 100

Step 1: Prime factorization of 100

Divisor Number
2 100
2 50
5 25
5 5
1

Step 2: Prime factorization expression

\[100 = 2 \times 2 \times 5 \times 5 = 2^2 \times 5^2\]

Step 3:

2 and 5 are without triplet

To make it a perfect cube, We need to multiply it by 2 × 5 = 10

∴ The smallest number is 10.

3. Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube.

(i) 81 (ii) 128 (iii) 135 (iv) 192 (v) 704

(i) 81

Step 1: Prime factorization of 81

Divisor Number
3 81
3 27
3 9
3 3
1

Step 2: Prime factorization expression

\[81 = 3 \times 3 \times 3 \times 3 = 3^3 \times 3 \]

Step 3: Determine what to divide

To make it a perfect cube, We need to divide it by 3.

∴ The smallest number is 3.

(ii) 128

Step 1: Prime factorization of 128

Divisor Number
2 128
2 64
2 32
2 16
2 8
2 4
2 2
1

Step 2: Prime factorization expression

\[128 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^3 \times 2^3 \times 2 \]

Step 3: Determine what to divide

To make it a perfect cube, We need to divide it by 2.

∴ The smallest number is 2.

(iii) 135

Step 1: Prime factorization of 135

Divisor Number
3 135
3 45
3 15
5 5
1

Step 2: Prime factorization expression

\[135 = 3 \times 3 \times 3 \times 5 = 3^3 \times 5 \]

Step 3: Determine what to divide

To make it a perfect cube, We need to divide it by 5.

∴ The smallest number is 5.

(iv) 192

Step 1: Prime factorization of 192

Divisor Number
2 192
2 96
2 48
2 24
2 12
2 6
3 3
1

Step 2: Prime factorization expression

\[192 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 2^6 \times 3 \]

Step 3: Determine what to divide

To make it a perfect cube, We need to divide by 3.

∴ The smallest number is 3.

(v) 704

Step 1: Prime factorization of 704

Divisor Number
2 704
2 352
2 176
2 88
2 44
2 22
11 11
1

Step 2: Prime factorization expression

\[704 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 11 = 2^6 \times 11 \]

Step 3: Determine what to divide

To make it a perfect cube, We need to divide it by 11.

∴ The smallest number is 11.

4. Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube?

Step 1: Find the volume of one cuboid

Volume = length × width × height = 5 cm × 2 cm × 5 cm = 50 cm³

Step 2: Find the prime factors of the volume

50 = 2 × 5 × 5 = 2¹ × 5²

Step 3: Determine what's needed to make it a perfect cube

To make a perfect cube, we need the exponents to be multiples of 3.

We need to multiply by 2² × 5¹ = 4 × 5 = 20

So the volume of the cube would be 50 × 20 = 1000 cm³

Step 4: Calculate how many cuboids are needed

Number of cuboids = Volume of cube ÷ Volume of one cuboid

= 1000 cm³ ÷ 50 cm³ = 20

∴ Parikshit will need 20 such cuboids to form a cube.