What is the remainder when a n-digit number 888…88 formed by 2020’s “8” is divided by 16?

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1. 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 $1\times 8 = 8$.

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