Question

Find the smallest number that should be multiplied with $7803$ so that the product becomes a perfect cube.

Answer 1

By prime factorising $7803$,

we get, $7803=3\times 3\times 3\times 17\times 17$

$=3^3\times {17}^2$

We know, a perfect cube has multiples of $3$ as powers of prime factors.

Here, power of $3$ is $3$ but power of $17$ is $2.$

So we need to multiply another $17$ to the factorization to make $7803$ a perfect cube.

Hence, the smallest number by which $7803$ must be multiplied to obtain a perfect cube is $17$.