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 $12-5=(\text{B}) 7$ factors have more than 3 factors. 
How many positive integer factors of 2020 have more than 3 factors? (As an example, 12 has 6 factors, namely $1,2,3,4,6$, and 12 .) (2020 AMC 8 Problem, Question \#17) 
Options:
  * $$6$$ 
  * $$7$$ 
  * $$8$$ 
  * $$9$$ 
  * $$10$$ 