Set
$A = \underbrace{8\dots8}_{2020 number 8's}$
We have
$\underbrace{8\dots8}_{2020 number 8's} = 8 \times \underbrace{1\dots1}_{2020 number 1's}$
Then
$A : 16 = (8 \times \underbrace{1\dots1}_{2020 number 1's}: 8):2$
$= \underbrace{1\dots1}_{2020 number 1's} : 2$
Observe that $ \underbrace{1\dots1}_{2020 number 1's}$ is an odd number, then the division above has the remainder of $1$.
Then the remainder when a n-digit number 888…88 formed by 2020’s “8” is divided by 16 is $1\times 8 = 8$.
