Find the least positive integer with $24$ positive divisors. 
Find the least positive integer with $24$ positive divisors.

My attempt:
$24=2^3.3$. We shall have to find out a positive integer (least) $n$ such that $N$ has $24$ positive divisors i.e we have to find $N$ where $\tau(N)=24.$.
We have, if $N=p_{1}^{\alpha_2}p_{2}^{\alpha_2}\dots p_{n}^{\alpha_n}$ then $\tau(N)=(\alpha_1+1)(\alpha_2+1)\dots (\alpha_n+1)$ where $p_i$, $i=1,2,\dots n$ are distinct primes and $\alpha_i$'s are all +ve.
Here $\tau(N)=(\alpha_1+1)(\alpha_2+1)\dots (\alpha_n+1)=24$. Now the problem is to find $\alpha_i$'s only.
The all possible factorizations are the following:
$24=1.24;~~ 24=2.12;~~ 24=3.8;~~ 24=4.6$. But how to get  $\alpha_i$'s?

We use the fact $p_1^{\alpha_1}p_2^{\alpha_2}\dots p_n^{\alpha_n}$ has $(\alpha_1+1)(\alpha_2+1)\dots(\alpha_n+1)$ divisors, and check minimum integer for each factorization, the possible factorizations for $24=2\cdot2\cdot2\cdot3$ are the following:
$2\cdot2\cdot2\cdot3$ minimum for this factorization is $2^2\cdot3\cdot5\cdot7=420$
$6\cdot2\cdot2$ minimum is $2^5\cdot3\cdot5=480$
$4\cdot2\cdot3$ minimum is $2^3\cdot3^2\cdot5=360$
$4\cdot 6$ minimum is $2^5\cdot3^3=864$
$12\cdot 2$ minimum is $2^{11}\cdot3>1000$
$8\cdot 3$ minimum is $2^{7}\cdot3^2>1000$
$24$ minimum is $2^{23}>1000000$
the minimum of the numbers in the right is $360$, which is our answer.
