Exercise 7.1

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

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

(v) 46656

Sol. (i) 216

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

Prime factors of 216 = 2 x 2 x 2 x 3 x 3 x 3

Here, each prime factor is appearing in multiple of 3.

Hence, 216 is a perfect cube.

(ii) 128

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

Prime factors of 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2

Here, each prime factor is not appearing in multiple of 3.

Hence, 128 is not a perfect cube.

(iii) 1000

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

Prime factors of 1000 = 2 x 2 x 2 x 5 x 5 x 5

Here, each prime factor is appearing in multiple of 3.

Hence, 1000 is a perfect cube.

(iv) 100

2	100
2	50
5	25
5	5
	1

Prime factors of 1000 = 2 x 2 x 5 x 5

Here, each prime factor is not appearing in multiple of 3.

Hence, 100 is not a perfect cube.

(v) 46656

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

Prime factors of 128 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3

Here, each prime factor is appearing in multiple of 3.

Hence, 46656 is a perfect cube.

Q.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) 6750

(v) 100

Sol. (i) 243

3	243
3	81
3	27
3	9
3	3
	1

Prime factors of 243 = 3 x 3 x 3 x 3 x 3

Here, prime factor ‘3’ is not appearing in triplet. Hence, to make 243 a cube, one more ‘3’ needs to be multiplied.

Therefore, 243 x 3 = 3 x 3 x 3 x 3 x 3 x 3 = 729 is a perfect cube.

Hence, the smallest number by which 243 must be multiplied to obtain a perfect cube is 3.

(ii) 256

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

Prime factors of 256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2

Here, prime factor ‘2’ is not appearing in triplet. Hence, to make 243 a cube, one more ‘2’ needs to be multiplied.

Therefore, 256 x 2 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 512 is a perfect cube.

Hence, the smallest number by which 256 must be multiplied to obtain a perfect cube is 2.

(iii) 72

2	72
2	36
2	18
3	9
3	3
	1

Prime factors of 72 = 2 x 2 x 2 x 3 x 3

Here, prime factor ‘3’ is not appearing in triplet. Hence, to make 72 a cube, one more ‘3’ needs to be multiplied.

Therefore, 72 x 3 = 2 x 2 x 2 x 3 x 3 x 3 = 216 is a perfect cube.

Hence, the smallest number by which 72 must be multiplied to obtain a perfect cube is 3.

(iv) 675

3	675
3	225
3	75
5	25
5	5
	1

Prime factors of 675 = 3 x 3 x 3 x 5 x 5

Here, prime factor ‘5’ is not appearing in triplet. Hence, to make 675 a cube, one more ‘5’ needs to be multiplied.

Therefore, 675 x 5 = 3 x 3 x 3 x 5 x 5 x 5 = 3375 is a perfect cube.

Hence, the smallest number by which 675 must be multiplied to obtain a perfect cube is 5.

(v) 100

2	100
2	50
5	25
5	5
	1

Prime factors of 100 = 2 x 2 x 5 x 5

Here, prime factor ‘2’ and ‘5’ are not appearing in triplet. Hence, to make 100 a cube, one more ‘2’ and ‘5’ needs to be multiplied.

Therefore, 100 x 2 x 5 = 2 x 2 x 2 x 5 x 5 x 5 = 1000 is a perfect cube.

Hence, the smallest number by which 100 must be multiplied to obtain a perfect cube is 2 x 5 = 10.

Q.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

Sol. (i) 81

3	81
3	27
3	9
3	3
	1

Prime factors of 81 = 3 x 3 x 3 x 3

Here, prime factor ‘3’ is not appearing in triplet. Hence, to make 243 a cube, one ‘3’ needs to be divided.

Therefore, 81 ÷ 3 = 3 x 3 x 3 = 27 is a perfect cube.

Hence, the smallest number by which 81 must be divided to obtain a perfect cube is 3.

(ii) 128

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

Prime factors of 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2

Here, prime factor ‘2’ is not appearing in triplet. Hence, to make 128 a cube, one ‘2’ needs to be divided.

Therefore, 128 ÷ 2 = 2 x 2 x 2 x 2 x 2 x 2 = 64 is a perfect cube.

Hence, the smallest number by which 128 must be divided to obtain a perfect cube is 2.

(iii) 135

3	135
3	45
3	15
5	5
	1

Prime factors of 135 = 3 x 3 x 3 x 5

Here, prime factor ‘5’ is not appearing in triplet. Hence, to make 135 a cube, one ‘5’ needs to be divided.

Therefore, 135 ÷ 5 = 3 x 3 x 3 = 27 is a perfect cube.

Hence, the smallest number by which 135 must be divided to obtain a perfect cube is 5.

(iv) 192

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

Prime factors of 192 = 2 x 2 x 2 x 2 x 2 x 2 x 3

Here, prime factor ‘3’ is not appearing in triplet. Hence, to make 192 a cube, one ‘3’ needs to be divided.

Therefore, 192 ÷ 3 = 2 x 2 x 2 x 2 x 2 x 2 = 64 is a perfect cube.

Hence, the smallest number by which 192 must be divided to obtain a perfect cube is 3.

(v) 704

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

Prime factors of 704 = 2 x 2 x 2 x 2 x 2 x 2 x 11

Here, prime factor ‘11’ is not appearing in triplet. Hence, to make 704 a cube, one ‘11’ needs to be divided.

Therefore, 704 ÷ 11 = 2 x 2 x 2 x 2 x 2 x 2 = 64 is a perfect cube.

Hence, the smallest number by which 704 must be divided to obtain a perfect cube is 11.

Q.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?

Sol. Given, sides of plasticine are 5 cm, 2 cm, 5 cm.

We know that, volume = 5 cm x 2 cm x 5 cm = (5 x 5 x 2) cm3.

Here, two 5s and one 2 are which are not in a triplet.

Hence, multiplying this by 2 x 2 x 5 = 20, then it will be a perfect cube.

Thus, (5 x 5 x 2 x 2 x 2 x 5) = (5 x 5 x 5 x 2 x 2 x 2) = 1000 is a perfect cube.

Hence, 20 cuboids will be required to form a cube.