To find solutions of $\dfrac n{d(n)}=p$ For positive integer $n$ let $d(n)$ denote the no, of positive divisors of $n$ , then for a prime $p$ , how do we find all solutions of $ \dfrac n{d(n)}=p$  ?
 A: Well, if what you want to know is how we find all solutions, there is a good notice: the set of solutions is bounded (the bound depends on $p$). I have found a very coarse bound: surely it can be lowered.
First, write
$$n=q_1^{\alpha_1}\cdots q_k^{\alpha_k}$$
Then, the equation is
$$p=\prod_j\frac{q_j^{\alpha_j}}{\alpha_j+1}$$
But each factor in this product is
$$\frac{q_j^{\alpha_j}}{\alpha_j+1}\geq 1$$
Then, none of them is greater than $p$. Moreover, the function
$$f(n,\alpha)=\frac{n^{\alpha}}{\alpha+1}$$
is strictly increasing and diverges to infinity for $n$ and for $\alpha$. Then, for a finitely many values of $\alpha$ there exists $q_\alpha$, the greatest prime such that
$$\frac{q_\alpha^{\alpha}}{\alpha+1}\leq p$$
For greater values of $\alpha$ simply there's no $q_\alpha$ thas satisfies the inequality.
The bound for the solution is
$$\prod_{\alpha}q_\alpha^\alpha$$
Note that this product is actually finite. Note also that $q_\alpha$ can be the same prime for different values of $\alpha$, and that in this case, only the greatest of these $\alpha$ is relevant.
Example: $p=37$. We find:
$$q_1=73$$
$$q_2=7$$
$$q_3=5$$
$$q_4=3$$
$$q_5=q_6=q_7=q_8=2$$
so the solutions of $n/d(n)=37$ are bounded by $73\cdot7^2\cdot 5^3\cdot 3^4\cdot 2^8$
