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What is the smallest number by which 588 must be multiplied so that the product is a perfect cube?

Competition practiceNumber System, HCF & LCM

A126
B42
C21
D294

Step-by-Step Solution

588 =

2^{2} \times 3 \times 7^{2}

. To make every exponent 3 we need one more 2, two more 3s and one more 7:

2 \times 9 \times 7 = 126

(product

74088 = 4 2^{3}

).

Answer: A) 126

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