Question

By which smallest number must the following number be divided so that the quotient is a perfect cube?

8640

Answer 1

On factorising 8640 into prime factors, we get:

\[8640 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5\]

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

 

\[8640 = \left\{ 2 \times 2 \times 2 \right\} \times \left\{ 2 \times 2 \times 2 \right\} \times \left\{ 3 \times 3 \times 3 \right\} \times 5\]

It is evident that the prime factors of 8640 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 8640 is a not perfect cube. However, if the number is divided by 5, the factors can be grouped into triples of equal factors such that no factor is left over.

Thus, 8640 should be divided by 5 to make it a perfect cube.