Japan 2007

For positive integer $n$, $$11^n=(10+1{)}^n=\sum _{k=0}^n{C}_n^{k}10^k=n\cdot 10^n+n\cdot 10^0=n\cdot 10^n+1(\mathrm{mod}100).$$ Therefore, the ten's place of $11^n$ is equal to the last digit of $n$. The last digit of $12^n$ for $n=1,2,\dots$ is $2,4,8,6,2,4,\dots$, with period $4$, so the last digit of $12^{13}$ is $2$ and the answer is $2$.