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?
 A: 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.
A: What usually helps me is expanding out the prime factorization of $24 = 2\cdot2\cdot2\cdot3$. We know that the exponents of the prime factorization of the number we need to find (call it $N$) must have exponents $1$ less than the actual values of numbers in the prime factorization of $24$. So the exponents should be $1, 1, 1, 2$. Let us let the number with exponent $2$ be the lowest prime number (we want to minimize $N$). So $N$ can equal $2^2\cdot3\cdot5\cdot7$. Note that we use $3, 5, 7$ because they are the three smallest primes after $2$. $$2^2\cdot3\cdot5\cdot7 = 420$$ That seems just a little too big does it not? Well you are probably thinking that $7$ must be too big of a prime number and you are indeed correct! Notice that if we get rid of the $7$ we only have $3\cdot2\cdot2 = 12$ factors. We need double that, so we can add $1$ to $2$ of the numbers' exponents we have so far obtained. Let us do it to the $2$ smallest of them (again we want to minimize $N$) We now have:
$$2^3\cdot3^2\cdot5 = 360$$
We can use the same logic as above and say $5$ is too big of a prime but if we get rid of it we'll still only have $12$ factors so we want to add $1$ to $2$ of the numbers' exponents (same explanation as when we removed the $7$). So we add one to the exponents of the smallest two numbers and get:$$2^4\cdot3^3 = 432$$But notice that $432$ is larger than $360$!
That means that the smallest value possible is $360$.
$$\boxed{N = 360}$$
A: For each $i$ we have $\alpha_i+1 \geq 2$. Therefore
$$2^n \leq (\alpha_1+1)(\alpha_2+1\dots (\alpha_n+1) \leq 24 \Rightarrow n \leq 4$$
Now discuss the four cases: $n=1, n=2, n=3$ and $n=4$.
