Q1: Which of the following numbers are not perfect cubes? (i) 216 (ii) 128 (iii) 1000 (iv) 100 (v) 46656
Sol:
(i) We have 216 = 2 × 2 × 2 × 3 × 3 × 3Grouping the prime factors of 216 into triples, no factor is left over.∴ 216 is a perfect cube.
(ii) We have 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2Grouping the prime factors of 128 into triples, we are left with 2 as an ungrouped factor.∴ 128 is not a perfect cube.
(iii) We have 1000 = 2 × 2 × 2 × 5 × 5 × 5Grouping the prime factors of 1000 into triples, we are not left with any factor.∴ 1000 is a perfect cube.
(iv) We have 100 = 2 × 2 × 5 × 5Grouping the prime factors into triples, we do not get any triples. Factors 2 ×2 and 5 ×5 are not triples.∴ 100 is not a perfect cube.
(v) We have 46656 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3Grouping the prime factors of 46656 in triples, we are not leftover with any prime factor.∴ 46656 is a perfect cube.
Q2: 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 Sol:
(i) We have 243 = 3 × 3 × 3 × 3 × 3The prime factor 3 is not a group of three.∴ 243 is not a perfect cube. Now, [243] × 3 = [3 × 3 × 3 × 3 × 3] × 3 or 729 = 3 × 3 × 3 × 3 × 3 × 3 Now, 729 becomes a perfect cube. Thus, the smallest required number to multiply 243 to make it a perfect cube is 3.
(ii) We have 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2Grouping the prime factors of 256 triples, we are left over with 2 × 2.∴ 256 is not a perfect cube. Now, [256] × 2 = [2 × 2 × 2 × 2 × 2 × 2 × 2 × 2] × 2 or 512 = 2 ×2 ×2 ×2 ×2 ×2 ×2 ×2 × 2 i.e. 512 is a perfect cube. Thus, the required smallest number is 2.
(iii) We have 72 = 2 × 2 × 2 × 3 × 3Grouping the prime factors of 72 in triples, we are left over with 3 ×3.∴ 72 is not a perfect cube. Now, [72] × 3 = [2 × 2 × 2 × 3 × 3] × 3 or 216 = 2 × 2 × 2 × 3 × 3 × 3 i.e. 216 is a perfect cube. ∴ The smallest number required to multiply 72 to make it a perfect cube is 3.
(iv) We have 675 = 3 × 3 × 3 × 5 × 5Grouping the prime factors of 675 to triples, we are left over with 5 × 5 675 is not a perfect cube. Now, [675] × 5 = [3 × 3 × 3 × 5 × 5] × 5 or 3375 = 3 × 3 × 3 × 5 × 5 × 5
Now, 3375 is a perfect cube.Thus, the smallest required number to multiply 675 such that the new number is a perfect cube is 5.
(v) We have 100 = 2 × 2 × 5 × 5The prime factor is not in the groups of triples.∴ 100 is not a perfect cube. Now, [100] × 2 × 5 = [2 × 2 × 5 × 5] × 2 × 5 or [100] ×10 = 2 × 2 × 2 × 5 × 5 × 5 1000 = 2 × 2 × 2 × 5 × 5 × 5 Now, 1000 is a perfect cube. Thus, the required smallest number is 10.
Q3: 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) We have 81 = 3 × 3 × 3 × 3Grouping the prime factors of 81 into triples, we are left with 3.∴ 81 is not a perfect cube. Now, [81] /3 = [3 × 3 × 3 × 3] ÷ 3 or 27 = 3 × 3 × 3 i.e. 27 is a perfect cube Thus, the required smallest number is 3.
(ii) We have 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2Grouping the prime factors of 128 into triples, we are left with 2.∴ 128 is not a perfect cube Now, [128] /2 = [2 × 2 × 2 × 2 × 2 × 2 × 2] ÷ 2 or 64 = 2 × 2 × 2 × 2 × 2 × 2 i.e. 64 is a perfect cube. ∴ The smallest required number is 2.
(iii) We have 135 = 3 × 3 × 3 × 5Grouping the prime factors of 135 into triples, we are left over with 5.∴ 135 is not a perfect cube Now, [135] /5 = [3 × 3 × 3 × 5] ÷ 5 or 27 = 3 × 3 × 3 i.e. 27 is a perfect cube. Thus, the required smallest number is 5.
(iv) We have 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3Grouping the prime factors of 192 into triples, 3 is left over.∴ 192 is not a perfect cube. Now, [192] / 3 = [2 × 2 × 2 × 2 × 2 × 2 × 3] ÷ 3 or 64 = 2 × 2 × 2 × 2 × 2 × 2 i.e. 64 is a perfect cube. Thus, the required smallest number is 3.
(v) We have 704 = 2 × 2 × 2 × 2 × 2 × 2 × 11Grouping the prime factors of 704 into triples, 11 is left over.∴ 704 is not a perfect cube. ∴ [704] /11 = [2 × 2 × 2 × 2 × 2 × 2 × 11] ÷ 11 or 64 = 2 × 2 × 2 × 2 × 2 × 2 i.e. 64 is a perfect cube. Thus, the required smallest number is 11.
Q4: 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: Sides of the cuboid are: 5 cm, 2 cm, 5 cm ∴ Volume of the cuboid = 5 cm × 2 cm × 5 cm To form it as a cube, its dimension should be in the group of triples.
Given numbers = 5 × 2 × 5
Since, Factors of 5 and 2 both are not in group of three. Therefore, the number must be multiplied by 2 × 2 × 5 = 20 to make it a perfect cube.
