Question

What is the least positive integer $n$ such that $\frac{{\left(30\right)}^3}{{\left(35\right)}^2}n$ is also integer?

Answer 1

$\frac{(30{)}^3}{(35{)}^2}n$

$=\frac{(2\times 3\times 5{)}^3}{(7\times 5{)}^2}n$

$=\frac{2^3\times 3^3\times 5^3}{7^2\times 5^2}n$

$=\frac{2^3\times 3^3\times 5^{3-2}}{7^2}n$

$=\frac{2^3\times 3^3\times 5}{7^2}n$

To get the result as integer, the denominator should be reduced to $1$ and that is possible only when we have $7^2$ in the numerator.

Since, $n$ is not given we can take it $7^2$ to get the final result as integer.

$\frac{2^3\times 3^3\times 5\times 7^2}{7^2}=2^3\times 3^3\times 5$ is an integer.

Thus, the value of $n=7^2=49$.