Find the highest natural number which is divisible by $30$ and has exactly $30$ different positive divisors. 
Find the highest natural number which is divisible by $30$ and have exactly $30$ different positive divisors.

What I Tried: I am not sure about any specific approach to this problem. Of course as the number is divisible by $30$ , some of the factors of the number would be $1,2,3,5,6,10,20,30$ , but that only makes $8$ divisors and there will be $12$ more which I have no idea of .
Can anyone help me?
 A: (I'm assuming you know the number theoretic way to count the number of divisors of an integer $ n = 2^a \times 3^b \times 5^c \ldots $, which is $ \sigma_0 (n) = (a+1)(b+1)(c+1) \ldots$. If not, go read up on it.)
Let $n$ be such a number.
Hint: Since $ 30 = 2 \times 3 \times 5 $, show that $ n = 2^a \times 3^b \times 5^c$.
In particular, conclude that $n$ cannot have any other prime factors.

 Note that every distinct prime factor of $n$ contributes at least one prime factor (not necessarily distinct) of $ \sigma_0 (n)$.
 Since $30 \mid n$, so $n$ has at least 3 prime factors (2, 3, 5), which contribute at least 3 (not necessarily distinct) prime factors to $ \sigma_0 (n) = 30$. Since this has 3 prime factors, we conclude that $n$ can have no other prime factors.
 Hence, $n = 2^a \times 3^b \times 5^c$.

Hence, the largest natural number is $ 5^4 \times 3^2 \times 2^1$.

Generlization: If $N = \prod p_i$ is the product of distinct primes, then the highest natural number which is divisible by $N$ and has exactly $N$ different positive divisors is $\prod p_i ^ { p_i - 1 } $.
A: If $n$ has $k$ prime divisors then $$\sigma (n) \geq \underbrace{(1+1)(1+1)\cdots (1+1)}_{k\; \rm times} = 2^k$$
So $2^k\leq 30$ and thus $k\leq 4$. So $n$ has $3$ or $4$ prime divisors.

*

*If $\color{red}{k=4}$ then $n= 2^a3^b5^cp^d$, where $p$ is prime greater than $5$. Now we have $$(a+1)(b+1)(c+1)(d+1)= 30$$ which is impossible.

*If $\color{green}{k=3}$ then $n= 2^a3^b5^c$. Now we have $$(a+1)(b+1)(c+1)= 30$$ which is possible if $a=1$, $b=2$ and $c=4$ or any permutation of them.

A: 
Not a 'real' answer, but it was too big for a comment. I think that you're looking for a solution without using a calculator or PC but maybe this gives some insight. I did only a quick search with the following bound: $1\le\text{n}\le10^8$.

I wrote and ran some Mathematica-code:
In[1]:=Clear["Global`*"];
\[Alpha] = 10^8;
ParallelTable[
  If[TrueQ[Length[Divisors[n]] == 30 && IntegerQ[n/30]], n, 
   Nothing], {n, 1, \[Alpha]}] //. {} -> Nothing

Running the code gives:
Out[1]={720, 1200, 1620, 4050, 7500, 11250}

We can see that for $1\le\text{n}\le10^8$ the number $\text{n}=11250$ is the biggest possible with the desired properties:
In[2]:=11250/30

Out[2]=375

In[3]:=Divisors[11250]

Out[3]={1, 2, 3, 5, 6, 9, 10, 15, 18, 25, 30, 45, 50, 75, 90, 125, 150, 225, 
250, 375, 450, 625, 750, 1125, 1250, 1875, 2250, 3750, 5625, 11250}

A: 30=2x3x5
So the exponents of 2,3, and 5 have to be 2,1,4 in no particular order. In order to maximize our product, we can do 5^4 x 3^2 x 2^1=11250
A: Because it must be a multiple of $30 = 2 \cdot 3 \cdot 5$, then the number must be of the form $$N = 2^{\alpha+1} \cdot 3^{\beta+1} \cdot 5^{\gamma+1} \cdot M $$ where $M$ is $1$ or a product of primes that are greater than $5$ and $\alpha, \beta$, and $\gamma$ are non negative integers.
Let $d(n)$ represent the number of different positive divisors of $n$. Then
$$d(N) = (\alpha+2)(\beta +2)(\gamma + 2)d(M)$$
and $d(N)$ is a multiple of $30 = 2 \cdot 3 \cdot 5$. It follows that
$\{\alpha+2, \beta+2, \gamma+2\} = \{2,3,5\}$ and $d(M)=1$. The largest value of $N$ will occur when $\alpha = 0, \beta = 1$, and $\gamma = 3$. The value of $N$ is therefore
$$N = 2^1 \cdot 3^2 \cdot 5^4 = 11250$$
