When the expression $2005^2 + 2005^0 + 2005^0 + 2005^5$ is evaluated, what are the final two digits?
That means I can ignore any multiples of 100 in each term, and just focus on the remainder when dividing by 100. For example, $2005^2 \equiv 5^2 \equiv 25 \pmod{100}$. Similarly, $2005^0 \equiv 1 \pmod{100}$ and $2005^5 \equiv 5^5 \pmod{100}$. So, the expression becomes $25 + 1 + 1 + 5^5 \pmod{100}$. To find $5^5 \pmod{100}$, I can use the fact that $5^4 \equiv 625 \equiv 25 \pmod{100}$, and multiply by another 5. That gives $5^5 \equiv 25 \times 5 \equiv 125 \equiv 25 \pmod{100}$. So, the expression becomes $25 + 1 + 1 + 25 \pmod{100}$. Adding these terms, I get $52 \pmod{100}$. Therefore, the final two digits of the original expression are 52. 