4. (10 points) A sequence of numbers: $8,3,1,4, . \cdots$, starting from the third number, each number is the unit digit of the sum of the two closest preceding numbers. What is the 2011th number? $\qquad$ .
【Analysis】Continuing to write down the sequence, we find that 60 numbers form a cycle. Then, 2011 divided by 60 gives 33 with a remainder of 31, meaning the 2011th number is the same as the 31st number. This can be used to solve the problem.

【Solution】Solution. Write down the first few numbers of the sequence:
$$
\begin{array}{l}
8,3,1,4,5,9,4,3,7,0,7,7,4,1,5,6,1,7,8,5,3,8,1,9,0,9, \\
9,8,7,5,2,7,9,6,5,1,6,7,3,0,3,3,6,9,5,4,9,3,2,5,7,2, \\
9,1,0,1,1,2,3,5,8,3, \cdots .
\end{array}
$$

The first number = the 61st number, the second number = the 62nd number, ... 60 is the cycle of the numbers.
$$
2011=33 \times 60+31,
$$

The 31st number is 2. Therefore, the 2011th number is 2.