Question

Find the smallest number by which $9000$ should be divided so that the quotient becomes a perfect cube?

• A. $3$
• B. $9$
• C. $27$
• D. $1$

Answer 1

The correct option is B $9$

Prime factorising $9000$, we get,

$9000=2\times 2\times 2\times 3\times 3\times 5\times 5\times 5$

$=2^3\times 3^2\times 5^3$

We know, a perfect cube has prime factors in powers of $3$.

Here, number of $2$'s is $3$, number of $3$'s is $2$ and number of $5$'s is $3$.

So we need to remove $3^2$ from the factorization to make $9000$ a perfect cube.

Hence, the smallest number by which $9000$ must be divided to obtain a perfect cube is $3^2=9$.

Therefore, option $B$ is correct.