Question

What is the smallest number by which 108 must be divided so that the quotient is a perfect cube?

• A. 8
• B. 4
• C. 12
• D. 15

Answer 1

The correct option is B 4

Prime factorisation of 108:
$\begin{array}{cc}2& 108\\ 2& 54\\ 3& 27\\ 3& 9\\ 3& 3\\ & 1\end{array}$

$108=2\times 2\times \underset{–}{3\times 3\times 3}$
As you can see that in the prime factorisation 108, its prime factor 2 is not in triplet, so we have to remove all prime factor of 2 those are not in triplet.
$⟹$ There are 2 such factors. So, 108 must be divided by $2\times 2=4$ to get the quotient as perfect cube.