What is the smallest number by which 1372 may be divided so that the quotient is a perfect cube also find the cube root of the resulting 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) 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

The prime factor 3 is not a group of three.∴ 243 is not a perfect cube.Now, [243] x 3 = [3 x 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

Thus, the smallest required number to multipkly 243 to make it a perfect cube is 3.(ii) We have 256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2

Grouping the prime factors of 256 in triples, we are left over with 2 x 2.∴ 256 is not a perfect cube.Now, [256] x 2 = [2 x 2 x 2 x 2 x 2 x 2 x 2 x 2] x 2or, 512 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2i.e. 512 is a perfect cube.thus, the required smallest number is 2.(iii) we have 72 = 2 x 2 x 2 x 3 x 3

Grouping the prime factors of 72 in triples, we are left over with 3 x 3∴ 72 is not a perfect cube.Now, [72] x 3 = [2 x 2 x 2 x 3 x 3] x 3or, 216 = 2 x 2 x 2 x 3 x 3 x 3i.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 x 3 x 3 x 5 x 5

Grouping the prime factors of 675 to triples, we are left over with 5 x 5

∴ 675 is not a perfect cube.Now, [675] x 5 = [3 x 3 x 3 x 5 x 5] x 5Now, 3375 is a perfect cubeThus, the smallest required number to multiply 675 such that the new number is a perfect cube is 5.(v) We have 100 = 2 x 2 x 5 x 5The prime factor are not in the groups of triples.

∴ 100 is not a perfect cube.Now, [100] x 2 x 5 = [2 x 2 x 5 x 5] x 2 x 5or, [100] x 10 = 2 x 2 x 2 x 5 x 5 x 51000 = 2 x 2 x 2 x 5 x 5 x 5Now, 1000 is a perfect cubeThus, the required smallest number is 10