Question

What is the smallest number by which the following numbers must be multiplied, so that the products are perfect cubes?

iv. $7803$

Answer 1

Step: Find the smallest number

Resolving $7803$ into prime factors:

$$\begin{array}{cc}3& 7803\\ 3& 2601\\ 3& 867\\ 17& 289\\ 17& 17\\ & 1\end{array}$$

$\Rightarrow 7803=\{3\times 3\times 3\}\times 17\times 17$

Clearly, on grouping the prime factors in triplets of equal factors, $17$ occurs only twice.

Thus, we need to multiply $7803$ with $17$ to make the product a perfect cube.

Hence, $17$ is the smallest required number.