What's the largest number of prime divisors that a number $n$ could have? A natural number $n$ has 20 divisors. What's the largest number of different prime divisors that a number $n$ could have?
What I have is  $$(\alpha_1+1)(\alpha_2+1)(\alpha_3+1)\cdots(\alpha_k+1)=20$$ where $$n=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}$$
But what's next?..
 A: For your case, the answer is $3$. 
In general, if a number has $k$ divisors, what's the maximum number of prime divisors that it could have? 
Every integer can be written as the product of different primes raised to some power greater than or equal to $1$, for example $x = {p_1}^{n_1} {p_2}^{n_2} \ldots {p_m}^{n_m}$, with all $n_j \geq 1$. The number of divisors is $k = (n_1 + 1)(n_2 + 1)\ldots(n_m + 1)$. The number of prime divisors is $m$. 
$m$ is largest if the $n_j$ are as small as possible: Take $k$, factor it into prime factors, and count the prime factors according to their multiplicity. For example, if a number $x$ has $k = 20 = 2 \cdot 2 \cdot 5$ divisors, it can have at most $3$ prime factors since $2 \cdot 2 \cdot 5$ is the product of three primes. Numbers with $20$ divisors and $3$ prime divisors are the numbers of the form $pqr^4$ where $p$, $q$ and $r$ are distinct primes. 
A: the number of factors in the expression $(\alpha_1+1)(\alpha_2+2)\dots(\alpha_k+1)=20$ is at most the number of prime factors of $20$ with multiplicity, which is $3$.
An example of such a factorization is $2\times2\times 5$ which yields the number $n=2\times 3\times 5^4$ for example.
