Find the smallest number which when multiplied with 3600 will make the product a perfect cube.

Solution:

On factorising 3600 into prime factors, we get:

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

On grouping the factors in triples of equal factors, we get:

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

It is evident that the prime factors of 3600 cannot be grouped into triples of equal factors such that no factor is left over.

Therefore, 3600 is not a perfect cube.

However, if the number is multiplied by $(2 \times 2 \times 3 \times 5=60)$, the factors can be grouped into triples of equal factors such that no factor is left over.

Hence, the number 3600 should be multiplied by 60 to make it a perfect cube.

Also, the product is given as:

$3600 \times 60=\{2 \times 2 \times 2\} \times 2 \times 3 \times 3 \times 5 \times 5 \times 60$

$\Rightarrow 216000=\{2 \times 2 \times 2\} \times 2 \times 3 \times 3 \times 5 \times 5 \times(2 \times 2 \times 3 \times 5)$

$\Rightarrow 216000=\{2 \times 2 \times 2\} \times\{2 \times 2 \times 2\} \times\{3 \times 3 \times 3\} \times\{5 \times 5 \times 5\}$

To get the cube root of the produce 216000, take one factor from each triple.

Cube root $=2 \times 2 \times 3 \times 5=60$

Hence, the required numbers are 60 and 60.