Solution: Let $L(d, n)$ be the last digit of a number ending in $d$ to the $n$th power. For $n \geq 1$, we know that $L(0, n)=0, L(1, n)=1, L(5, n)=5, L(6, n)=6$. All numbers ending in odd digits in this series are raised to odd powers; for odd $n, L(3, n)=3$ or 7 , $L(7, n)=3$ or $7, L(9, n)=9$. All numbers ending in even digits are raised to even powers; for even $n, L(2, n)=4$ or $6, L(4, n)=L(6, n)=6, L(8, n)=6$ or 4 . Further, for each last digit that has two possible values, the possible values will be present equally as often. Now define $S(d)$ such that $S(0)=0$ and for $1 \leq d \leq 9, S(d)=L(d, d)+L(d, d+10)+$ $L(d, d+20)+L(d, d+30)+\cdots+L(d, d+90)$, so that the sum we want to calculate becomes $S(0)+S(1)+S(2)+\cdots+S(9)$. But by the above calculations all $S(d)$ are divisible by 10 , so their sum is divisible by 10 , which means its last digit is 0 .
6. What is the last digit of $1^{1}+2^{2}+3^{3}+\cdots+100^{100}$ ?