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) 704
NCERT 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:
(i) We have,
1536 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3
After grouping the prime factors in triplets, it’s seen that one factor 3 is left without grouping.
1536 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × 3
So, in order to make it a perfect cube, it must be divided by 3.
Thus, the smallest number by which 1536 must be divided to obtain a perfect cube is 3.
(ii) We have,
10985 = 5 × 13 × 13 × 13
After grouping the prime factors in triplet, it’s seen that one factor 5 is left without grouping.
10985 = 5 × (13 × 13 × 13)
So, it must be divided by 5 in order to get a perfect cube.
Thus, the required smallest number is 5.
(iii) We have,
28672 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 7
After grouping the prime factors in triplets, it’s seen that one factor 7 is left without grouping.
28672 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × 7
So, it must be divided by 7 in order to get a perfect cube.
Thus, the required smallest number is 7.
(iv) 13718 = 2 × 19 × 19 × 19
After grouping the prime factors in triplets, it’s seen that one factor 2 is left without grouping.
13718 = 2 × (19 × 19 × 19)
So, it must be divided by 2 in order to get a perfect cube.
Thus, the required smallest number is 2.
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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) We have 243 = 3 x 3 x 3 x 3 x 3
or, 729, = 3 x 3 x 3 x 3 x 3 x 3
Now, 729 becomes a [perfect cube
∴ The smallest number required to multiply 72 to make it a perfect cube is 3.
(iv) We have 675 = 3 x 3 x 3 x 5 x 5Grouping the prime factors of 675 to triples, we are left over with 5 x 5