5. (10 points) An electronic clock always displays the date as an eight-digit number, such as 20110101 for January 1, 2011. In 2011, the last day that can be divided by 101 is $\overline{2011 \mathrm{ABCD}}$, so $\overline{\mathrm{ABCD}}=$ $\qquad$ .
【Solution】Solution: First, analyze the divisibility characteristic of 101, which is that the difference between the odd segment and the even segment after two-digit truncation can be divided by 101.

Since the last day, let's see if there is one in December, let $\overline{\mathrm{AB}}=12$.
The sum of the even segments is $20+12=32$, so the sum of the odd segments must also be 32 to meet the condition, $32-11=21$ which means $\overline{\mathrm{CD}}=21$ $\overline{\mathrm{ABCD}}=1221$.

Method two: Trial division, let $\overline{\mathrm{AB}}=12$.
$$
20111299 \div 101=199121 \cdots 78.20111299-78=20111221 . \overline{\mathrm{ABCD}}=1221
$$

Therefore, the answer is: 1221.