Problem 688

For positive integer $n$,
$$11^n = (10 + 1)^n = \sum_{k=0}^{n} C_n^k 10^k = n \cdot 10^n + n \cdot 10^0 = n \cdot 10^n + 1 \pmod{100}.$$
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 $2$.