Notice that $1^{2008} = 1$. $2^1 = 2$, $2^2 = 4$, $2^3 = 8$, $2^4 = 16$, $2^5 = 32$, ... From this we conclude that all the powers of $2$ are ending on $2$, $4$, $8$, $6$ and they are repeating in that order, depending on the residuum of the power $1$, $2$, $3$ or $0$ when divided with $4$. Because $2008$ is divisible with $4$ we conclude that $2^{2008}$ ends in $6$. Similarly $3^{2008}$ ends in $1$, $4^{2008}$ ends in $6$, $5^{2008}$ ends in $5$ and $6^{2008}$ ends in $6$. Because $1+6+1+6+5+6=25$, which means that it ends on $5$, it follows that $1^{2008} + 2^{2008} + 3^{2008} + 4^{2008} + 5^{2008} + 6^{2008}$ ends on $5$, hence is divisible with $5$.