Prime Factors of 2016

$2016 = 2^5 \times 3^2 \times 7^1$ . Observing that we need all powers to be even in order for it to be a square number, we multiply by $2\times 7=14$ .

√2016=12√14 so,12×〖14〗^2 will be the perfect square number.the least positive integer is 14

We know that $\sqrt{k \times 2016} = \sqrt{a^{2}}$ . LHS and RHS are both Natural Numbers. K is the smallest integer we are looking for. Then $a = 4 \times 3 \times \sqrt{k} \times \sqrt{2 \times 7}.$ For k = 14 $a$ is a Natural Number. Since Prime factorization of $14 = 2 \times 7$ , there is no smaller k.

sq(2016) = 12 * sq(7) * sq(2) So, all you have to do is add a root 2 and root 7 to 2016, which means 2016 * 7 * 2 = 28224, then divide 28224 by 2016 because 2016 * x = 28224, and x = 14.