1. What is the hundreds digit of $5^{2017}$?
(A) 5
(B) 2
(C) 1
(D) 3
(E) 6
(1) The correct answer is $(C)$.

It is easily seen that $5 \cdot(1000 k+125)=5000 k+625$ and that $5 \cdot(1000 k+625)=5000 k+3125=$ $1000(5 k+3)+125$. Since $5^{3}=125$, the powers with odd exponents of 5, from the third onward, will end with the digits 125, while those with even exponents will end with 625. We can conclude that the hundreds digit of $5^{2017}$ is 1.

Problem proposed by Antonio Fanelli.