How many positive divisors does $20$ have?
As the prime factorization of $20$ is $2^2\times 5$. There are three ways to pick up $2$, including no $2$ is chosen. Similarly, there are $2$ ways to pick up $5$. Hence the result is $3\times 2=\boxed{6}$.
In a more concise way, because $20=2^2\times 5$, we find the answer is $(2+1)\times(1+1)=6$.