1. Is the sum $1^{2008}+2^{2008}+3^{2008}+4^{2008}+5^{2008}+6^{2008}$ divisible by 5? Explain your answer.
Solution. First, let's note that $1^{2008}=1$. Since $2^{1}=2$, $2^{2}=4,2^{3}=8,2^{4}=16,2^{5}=32, \ldots$, we conclude that the powers of the number 2 end in $2,4,8$ and 6 and repeat in this order depending on whether the exponent modulo 4 gives a remainder of $1,2,3$ or is divisible by 4. Since 2008 is divisible by 4, we have 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. Since $1+6+1+6+5+6=25$, which ends in 5, it follows that the sum $1^{2008}+2^{2008}+3^{2008}+4^{2008}+5^{2008}+6^{2008}$ also ends in 5, i.e., is divisible by 5.