What is the smallest number by which 8192 must be divided so that quotient is a perfect cube?

On factorising 8192 into prime factors, we get:

$8192=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$

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

$8192=\{2 \times 2 \times 2\} \times\{2 \times 2 \times 2\} \times\{2 \times 2 \times 2\} \times\{2 \times 2 \times 2\} \times 2$

It is evident that the prime factors of 8192 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 8192 is not a perfect cube. However, if the number is divided by 2, the factors can be grouped into triples of equal factors such that no factor is left over.

Hence, the number 8192 should be divided by 2 to make it a perfect cube.

Also, the quotient is given as:

$\frac{8192}{2}=\frac{\{2 \times 2 \times 2\} \times\{2 \times 2 \times 2\} \times\{2 \times 2 \times 2\} \times\{2 \times 2 \times 2\} \times 2}{2}$

$\Rightarrow 4096=\{2 \times 2 \times 2\} \times\{2 \times 2 \times 2\} \times\{2 \times 2 \times 2\} \times\{2 \times 2 \times 2\}$

To get the cube root of the quotient 4096, take one factor from each triple. We get: 

Cube root $=2 \times 2 \times 2 \times 2=16$

Hence, the required numbers are 2 and 16.