2017 AMC 10B Problem 14

By Fermat’s Little Theorem, $N^4\equiv1\pmod5$ whenever $N$ is not divisible by $5.$ Therefore, $$N^{16}\equiv(N^4)^4\equiv1\pmod5.$$ There are $\frac{2020}{5}=404$ multiples of $5,$ so there are $2020-404=1616$ allowable values of $N.$ A multiple of $5$ has $N^{16}\equiv0\pmod5,$ so no other values work. Thus the probability is $$\dfrac{1616}{2020}=\dfrac45.$$ Thus, the correct answer is D .