What is the remainder when $3^0 + 3^1 + 3^2 + \cdots + 3^{2009}$ is divided by 8?
The sum of any four consecutive powers of 3 is divisible by $3^0 + 3^1 + 3^2 +3^3 = 40$ and hence is divisible by 8. Therefore $(3^2 + 3^3 + 3^4 + 3^5) + \cdots + (3^{2006} + 3^{2007} + 3^{2008} + 3^{2009})$ is divisible by 8.  So the required remainder is $3^0 + 3^1 = \fbox{4}$. The answer is $\mathrm{(D)}$.