What is the remainder when $2^{2005}$ is divided by 7?
I know that $2^3 \equiv 1 \pmod{7}$, so I can use that to simplify the exponent. I can write $2005 = 3 \times 668 + 1$, so $2^{2005} = 2^{3 \times 668 + 1} = (2^3)^{668} \times 2^1$. Using the congruence, I get $(2^3)^{668} \times 2^1 \equiv 1^{668} \times 2^1 \equiv 2 \pmod{7}$. So the remainder is 2. 