10. Determine the units digit of $(2!+2)(3!+3)(4!+2)(5!+3) \cdots(2018!+2)(2019!+3)$.
Answer: 8
Solution: In mod 10, the expression above becomes $(4)(9)(6)\left(2^{1007}\right)\left(3^{1008}\right)=2^{1010} \cdot 3^{1011}$ because starting from 5!, the units digit will be alternating from 3 and 2. In $\bmod 10,2^{1010} \cdot 3^{1011}$ becomes $(4)(7)=28$, which ends with 8 .