Twenty divisors

If $n= {p_1}^{e_1}{p_2}^{e_2}\dots$ is the prime factorization of $n$ , then the number of divisors of $n$ is the product of the one more than the exponents $(e_1+1)(e_2+1)\dots$ . Since we are looking for a number with $20$ divisors, the product must be $20$ . The possible products for $20$ are $20, (2\times 10),( 2\times2\times5), \text{ and } (4\times 5$ ).

In the first case, the exponent is $19$ , so the smallest number with $20$ divisors with only one prime factor (which is chosen to be $2$ inorder to produce the smallest number) is $2^{19}$ .

Similarly, if the number has two prime factors, then the smallest values are $(3^1\times 2^9),( 3^3\times 2^4 )$ ( in these two cases, we choose $2$ and $3$ as the prime divisors to make the number as small as possible). Finally, if the number has $3$ prime factors, then the smallest value is $5^1\times3^1\times2^4 = \boxed{240}.$