What is the remainder when $3^0 + 3^1 + 3^2 + \cdots + 3^{2009}$ is divided by 8?
$\mathrm{(A)}\ 0\qquad \mathrm{(B)}\ 1\qquad \mathrm{(C)}\ 2\qquad \mathrm{(D)}\ 4\qquad \mathrm{(E)}\ 6$
Solution 1
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 $3^0 + 3^1 = \boxed {4}$. The answer is $\mathrm{(D)}$.

Solution 2
We have $3^2 = 9 \equiv 1 \pmod 8$. Hence for any $k$ we have $3^{2k}\equiv 1^k = 1 \pmod 8$, and then $3^{2k+1} = 3\cdot 3^{2k} \equiv 3\cdot 1 = 3  \pmod 8$.
Therefore our sum gives the same remainder modulo $8$ as $1 + 3 + 1 + 3 + 1 + \cdots + 1 + 3$. There are $2010$ terms in the sum, hence there are $2010/2 = 1005$ pairs of $1+3$, and thus the sum is $1005 \cdot 4 = 4020 \equiv 20 \equiv \boxed{4} \pmod 8$.

Solution 3
We have the formula $\frac{a(r^n-1)}{r-1}$ for the sum of a finite geometric sequence which we want to find the residue modulo 8.
\[\frac{1 \cdot (3^{2010}-1)}{2}\]
\[\frac{3^{2010}-1}{2} = \frac{9^{1005}-1}{2}\]
\[\frac{9^{1005}-1}{2} \equiv \frac{1^{1005}-1}{2} \equiv \frac{0}{2} \pmod 8\]
Therefore, the numerator of the fraction is divisible by $8$. However, when we divide the numerator by $2$, we get a remainder of $4$ modulo $8$, giving us $\mathrm{(D)}$.
Note: you need to prove that \[\frac{9^{1005}-1}{2}\] is not congruent to 0 mod 16 because if so, then the whole thing would be congruent to 0 mod 8, even after dividing by 2 ~ ilikepi12