Question

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

243000

Answer 1

On factorising 243000 into prime factors, we get:

\[243000 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 5 \times 5 \times 5\]

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

\[243000 = \left\{ 2 \times 2 \times 2 \right\} \times \left\{ 3 \times 3 \times 3 \right\} \times 3 \times 3 \times \left\{ 5 \times 5 \times 5 \right\}\]

It is evident that the prime factors of 243000 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 243000 is a not perfect cube. However, if the number is divided by (\[3 \times 3 = 9\]), the factors can be grouped into triples of equal factors such that no factor is left over.

Thus, 243000 should be divided by 9 to make it a perfect cube.