Question

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

7803

Answer 1

On factorising 7803 into prime factors, we get:

\[7803 = 3 \times 3 \times 3 \times 17 \times 17\]

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

\[7803 = \left\{ 3 \times 3 \times 3 \right\} \times 17 \times 17\]

It is evident that the prime factors of 7803 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 7803 is a not perfect cube. However, if the number is divided by (\[17 \times 17 = 289\]), the factors can be grouped into triples of equal factors such that no factor is left over.

 

Thus, 7803 should be divided by 289 to  make it a perfect cube.