Since $2020=2^{2} \cdot 5 \cdot 101$, we can simply list its factors:  $$ 1,2,4,5,10,20,101,202,404,505,1010,2020 . $$  There are $12$ of these; only $1,2,4,5,101$ (i.e. $5$ of them) don\textquotesingle t have over $3$ factors, so the remaining $12-5=(\mathbf{B}) 7$ factors have more than $3$ factors. 
How many positive integer factors of $2020$ have more than $3$ factors? (2020 AMC 8 Problems, Question \#17) 

1. $$6$$ 
2. $$7$$ 
3. $$8$$ 
4. $$9$$ 
5. $$10$$ 