Question

Find the smallest number by which $128$ must be divided, so that the quotient is a perfect cube.

• A. $2$
• B. $3$
• C. $7$
• D. $12$

Answer 1

Answer: (A)

$2$

$128=2\times 2\times 2\times 2\times 2\times 2\times 2$
$=(2^3\times 2^3\times 2)$
$={(2\times 2)}^3\times 2$
In the above factorization there is no triplet for $2$.
So, $128$ is not a perfect cube.
Therefore, $128$ must be divided by $2$ to make the quotient a perfect cube.