Question

By what least number must $3600$ be divided to make it a perfect cube?

• A. $9$
• B. $50$
• C. $300$
• D. $450$

Answer 1

Answer: (D)

$450$

Prime factorising $3600$, we get,
$3600=36\times 100$

$3600=6^2\times {10}^2$

$3600=(2\times 3{)}^2\times (2\times 5{)}^2$

$3600=3\times 3\times 2\times 2\times 2\times 2\times 5\times 5$

$3600=2^4\times 3^2\times 5^2$.

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

Here, number of $2$'s is $4$, number of $3$'s is $2$ and number of $5$'s is $2$.

So we need to divide $2$, $3^2$ and $5^2$ from the factorization to make $3600$ a perfect cube.

Hence, the smallest number by which $3600$ must be divided to obtain a perfect cube is $2\times 3^2\times 5^2=450$.

Therefore, option $D$ is correct.