What is the least number by which 135 may be divided so that the quotient is a perfect cube?

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Answer

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Hint: We will do prime factorization of 135.Further we will divide 135 to a number, that number is not in triplet form. Then we will convert 135 into a perfect cube.

Complete step by step solution:

Here the given number is $135$We will use prime factorization method

$3$	$135$
$3$	$45$
$3$	$15$
$5$	$5$
	$1$

$135 = 3 \times 3 \times 3 \times 5$$135 = {3^3} \times 5$Here, one $5$ is left which does not form a tripletIf we divide $135$ by $5$, then it will become a perfect cube.So, $\dfrac{{135}}{5} = {3^3}$$27 = {3^3}$Hence, the smallest number by which $135$ should be divided to make it a perfect cube is $5$.

Note: In these types of questions usually students get puzzled whether to find HCF or LCM. We note that words like larger, highest etc. are keywords mentioned in the question and they give us ideas to find HCF whereas words like smallest, lowest, least etc. give us direction to find the LCM.

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

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