2. (5 points) It is stipulated: the product of $n$ $a$s is denoted as: $\underbrace{\mathrm{a} \times \mathrm{a} \times \cdots \times \mathrm{a} \times \bar{c}}_{\mathrm{n} \text { a's }}=a^{n}$
Observe the following calculations: $8^{1}=8 ; 8^{2}=64 ; 8^{3}=512 ; 8^{4}=4096 ; 8^{5}=32768 ; 8^{6}=262144 ; 8^{7}=2097152$; $8^{8}=16777216, \cdots$
The remainder when $8^{2012}$ is divided by 10 is $\qquad$
【Analysis】Observing “$8^{1}=8 ; 8^{2}=64 ; 8^{3}=512 ; 8^{4}=4096 ; 8^{5}=32768 ; 8^{6}=262144 ; 8^{7}=2097152$; $8^{8}=16777216, \cdots$” we know that, except for the first term, the remainder of the subsequent terms when divided by 10 is $4, 2, 6, 8, 4, 2, 6, 8, \cdots$; that is, they cycle in a group of 4, 2, 6, 8. Therefore, to find the answer, we need to divide 2012-1 by 4 to determine how many cycles and the remainder, which will give us the answer.

【Solution】Solving: $(2012-1) \div 4$,
$$
\begin{array}{l}
=2011 \div 4, \\
=502 \cdots 3,
\end{array}
$$

This means the 3rd number in a cycle, so $8^{2012}$ divided by 10 gives a remainder of 6. Therefore, the answer is: 6.