TITLE: Smallest number with specific number of divisors
QUESTION [17 upvotes]: Is there a general method for finding smallest number of specific number of divisors? I am doing "Higher Algebra by Barnard JM Child" and came across a question that "find the smallest number with 24 divisors", that's how I tried to solve it, alert me if I am wrong:
Since $24$ can not have more than $4$ prime factors, the number can not have more than 4 prime factors.
As a single number it is :$2$^${23}$
as product of two numbers: $2^5*3^3$,$2^{11}*3$,$2^7*3^2$,
since $5+3<7+2<11+1$, so $2^5*3^3$ is the min of these numbers
as product of three numbers :$2^5*3*5$,$2^3*3^2*5$ since $3+2+1<5+1+1$, hence $2^3*3^2*5$ is the lesser of two
as product of 4 numbers $2^2*3*5*7$
$k=\min(2^{23},2^5*3^3,2^3*3^2*5,2^2*3*5*7)$
=$2^3*3^2*5=360$
The above method seems to be fishy and laborious, is there a general approach to find the smallest number with specific number of divisors?

REPLY [4 votes]: A005179 lists some resources on the problem. In particular:
Grost, M. (1968). The Smallest Number with a Given Number of Divisors. The American Mathematical Monthly, 75(7), 725-729. doi:10.2307/2315183
From the introduction:
Given $h = q_1 q_2 \dots q_n$, with primes $q_1 \ge q_2 \ge \dots \ge q_n$, let $A(h)$ be the smallest number with $h$ divisors.
In many cases, $$A(h) = 2^{q_1-1} 3^{q_2-1} \dots p_n^{q_n-1}$$
The primary objective of the paper is to determine the exceptions.
We call these numbers ordinary, from R. Brown, The minimal number with a given number of divisors, Journal of Number Theory 116 (2006) 150-158. In this paper Brown shows almost all $A(h)$ are ordinary, in particular "We show here that all square-free numbers are ordinary and that the set of ordinary numbers has natural density one."