<s>[INST] 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?
 [/INST] 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
$$N = 2^1 \cdot 3^2 \cdot 5^4 = 11250$$
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