Question

By which smallest number must the following numbers be divided so that the quotient is a perfect cube?
$\left(v\right)7803$

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, we observe that the factor $17$ is not forming a triplet.

Thus, we need to divide $7803$ by $17\times 17=289$ to make the quotient a perfect cube.

Hence, $289$ is the smallest required number.