$\tau (x) = 2016$ and for least $x = a \times b^2$ find $b$.

Question

$\tau(x) = 2016$. For the least value $x$ such that $x = a \times b^2$ and $a$ has no square divisors in it, find the value of $b$.
[ $\tau (n) = \text{number of divisors of $n$}.$

This seems a quite easy problem. If $a$ and $b$ was prime then I would have solved it immediately. But they are not prime. So, I can't find a way to start with.

Source: BdMO 2016 question set 5 problem 8 (Higher Secondary).

Answer 1

$$x=p_1p_2p_3\cdots p_nq_1^{2\alpha_1}q_2^{2\alpha}\cdots q_m^{2\alpha_m}$$ $$\tau(x)=(1+1)^n(2\alpha_1+1)(2\alpha_2+1)\cdots (2\alpha_m+1)$$ Since $2016=2^5\cdot3^2\cdot7$ we have $$n=5,\space \alpha_1=1,\space\alpha_2=1,\space \alpha_3=3$$

All number of the form $x=p_1p_2p_3p_4p_5 q_1^2q_2^2q_3^6$ where the primes are distinct satisfies $\tau(x)=2016$.

If we want to have a minimum $x$ we must have $$x=2^6\cdot3^2\cdot5^2\cdot 7\cdot 11\cdot 13\cdot 17\cdot 19$$ The corresponding value of $b$ is $$b=2^3\cdot3\cdot5=\color{red}{120}$$