No worries! We‘ve got your back. Try BYJU‘S free classes today! Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses No worries! We‘ve got your back. Try BYJU‘S free classes today! No worries! We‘ve got your back. Try BYJU‘S free classes today!
(i)First find the prime factors of 675 675 = 3 × 3 × 3 × 5 × 5 = 33 × 52 Since 675 is not a perfect cube. To make the quotient a perfect cube we divide it by 52 = 25, which gives 27 as quotient where, 27 is a perfect cube. ∴ 25 is the required smallest number. (ii) 8640 First find the prime factors of 8640 8640 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5 = 23 × 23 × 33 × 5 Since 8640 is not a perfect cube. To make the quotient a perfect cube we divide it by 5, which gives 1728 as quotient and we know that 1728 is a perfect cube. ∴5 is the required smallest number. (iii) 1600 First find the prime factors of 1600 1600 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 = 23 × 23 × 52 Since 1600 is not a perfect cube. To make the quotient a perfect cube we divide it by 52 = 25, which gives 64 as quotient and we know that 64 is a perfect cube ∴ 25 is the required smallest number. (iv) 8788 First find the prime factors of 8788 8788 = 2 × 2 × 13 × 13 × 13 = 22 × 133 Since 8788 is not a perfect cube. To make the quotient a perfect cube we divide it by 4, which gives 2197 as quotient and we know that 2197 is a perfect cube ∴ 4 is the required smallest number. (v) 7803 First find the prime factors of 7803 7803 = 3 × 3 × 3 × 17 × 17 = 33 × 172 Since 7803 is not a perfect cube. To make the quotient a perfect cube we divide it by 172 = 289, which gives 27 as quotient and we know that 27 is a perfect cube ∴ 289 is the required smallest number. (vi) 107811 First find the prime factors of 107811 107811 = 3 × 3 × 3 × 3 × 11 × 11 × 11 = 33 × 113 × 3 Since 107811 is not a perfect cube. To make the quotient a perfect cube we divide it by 3, which gives 35937 as quotient and we know that 35937 is a perfect cube. ∴ 3 is the required smallest number. (vii) 35721 First find the prime factors of 35721 35721 = 3 × 3 × 3 × 3 × 3 × 3 × 7 × 7 = 33 × 33 × 72 Since 35721 is not a perfect cube. To make the quotient a perfect cube we divide it by 72 = 49, which gives 729 as quotient and we know that 729 is a perfect cube ∴ 49 is the required smallest number. (viii) 243000 First find the prime factors of 243000 243000 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 5 × 5 × 5 = 23 × 33 × 53 × 32 Since 243000 is not a perfect cube. To make the quotient a perfect cube we divide it by 32 = 9, which gives 27000 as quotient and we know that 27000 is a perfect cube ∴ 9 is the required smallest number.
Solution: A number is a perfect cube only when each factor in the prime factorization is grouped in triples. Using this concept, the smallest number can be identified. (i) 81 81 = 3 × 3 × 3 × 3 = 33 × 3 Here, the prime factor 3 is not grouped as a triplet. Hence, we divide 81 by 3, so that the obtained number becomes a perfect cube. Thus, 81 ÷ 3 = 27 = 33 is a perfect cube. Hence the smallest number by which 81 should be divided to make a perfect cube is 3. (ii) 128 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2 = 23 × 23 × 2 Here, the prime factor 2 is not grouped as a triplet. Hence, we divide 128 by 2, so that the obtained number becomes a perfect cube. Thus, 128 ÷ 2 = 64 = 43 is a perfect cube. Hence the smallest number by which 128 should be divided to make a perfect cube is 2. (iii) 135 135 = 3 × 3 × 3 × 5 = 33 × 5 Here, the prime factor 5 is not a triplet. Hence, we divide 135 by 5, so that the obtained number becomes a perfect cube. 135 ÷ 5 = 27 = 33 is a perfect cube. Hence the smallest number by which 135 should be divided to make a perfect cube is 5. (iv) 192 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3 = 23 × 23 × 3 Here, the prime factor 3 is not grouped as a triplet. Hence, we divide 192 by 3, so that the obtained number becomes a perfect cube. 192 ÷ 3 = 64 = 43 is a perfect cube Hence the smallest number by which 192 should be divided to make a perfect cube is 3. (v) 704 704 = 2 × 2 × 2 × 2 × 2 × 2 × 11 = 23 × 23 × 11 Here, the prime factor 11 is not grouped as a triplet. Hence, we divide 704 by 11, so that the obtained number becomes a perfect cube. Thus, 704 ÷ 11 = 64 = 43 is a perfect cube Hence the smallest number by which 704 should be divided to make a perfect cube is 11. ☛ Check: NCERT Solutions for Class 8 Maths Chapter 7 Video Solution: 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) 704NCERT Solutions for Class 8 Maths Chapter 7 Exercise 7.1 Question 3 Summary: 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 are (i) 3, (ii) 2, (iii) 5, (iv) 3, and (v) 11 ☛ Related Questions:
Home » Aptitude » Simplification » Question
243000 = 243 × 1000= 3 × 3 × 3 × 3 × 3 × 10 × 10 × 10 = 33 × 32 × 103 ∴ Required number = 32 = 9
|