9. There are five honest security guards $A, B, C, D, E$ who each worked three days on December 1, 2016. Every day, 3 security guards were on duty, and each guard's schedule repeated every 5 days. Today (January 1, 2017, Sunday), regarding their duty schedule last month, the five had the following conversation:
$A$: $B$ and I worked more weekend (Saturday, Sunday) shifts than the other three.
$B$: I worked with each of the other four at least once this month.
$C$: On December 3, I was supposed to be off, but I came to help with the Math Garden Exploration Preliminary Contest, which unfortunately doesn't count as a duty shift.
$D$: $E$ is always scheduled with me.
$E$: On Christmas (December 25), both $A$ and I were on duty.
So, the five-digit number formed by the dates of the 2nd, 6th, and 10th times $A$ was on duty in December is $\qquad$.
(If the 2nd, 6th, and 10th times on duty were on December 3 and December 17, respectively, the answer would be 31217)
【Solution】Solution: The on-duty schedule for December is as follows:
\begin{tabular}{|c|c|c|c|c|c|}
\hline & Dec 1 (Mon) & 2 (Tue) & 3 (Wed) & 4 (Sun) & 5 (Fri) \\
\hline A & $\sqrt{ }$ & & & $\sqrt{ }$ & $\sqrt{ }$ \\
\hline B & Half & & $\checkmark$ & $\checkmark$ & \\
\hline C & $\checkmark$ & $\checkmark$ & & $\checkmark$ & \\
\hline D & & $\checkmark$ & $\checkmark$ & & $\checkmark$ \\
\hline E & & $\checkmark$ & $\sqrt{ }$ & & $\checkmark$ \\
\hline & 6 (Mon) & 7 (Tue) & 8 (Wed) & 9 (Thu) & 10 (Fri) \\
\hline & 11 (Sun) & 12 (Sat) & 13 (Mon) & 14 (Tue) & 15 (Sun) \\
\hline & 16 (Thu) & 17 (Fri) & 18 (Sat) & 19 (Sun) & 20 (Mon) \\
\hline & 21 (Tue) & 22 (Wed) & 23 (Thu) & 24 (Fri) & 25 (Sun) \\
\hline & 26 (Sat) & 27 (Mon) & 28 (Tue) & 29 (Wed) & 30 (Thu) \\
\hline & 31 (Fri) & & & & \\
\hline Weekend days & 2 & 1 & 2 & 2 & 2 \\
\hline
\end{tabular}

From what E said, we know that both A and E are on duty on the 25th. From what D said, we know that D and E are always together, so we can conclude that the people on duty on the 5th are A, D, and E.

From what C said, we know that C is not on duty on the 3rd. Since there must be 3 people on duty every day, one of D or E must be on duty, and since D and E are always together, both D and E are on duty on the 3rd.

From what A said, A and B are on duty more weekends than C, D, and E. Counting the number of weekends in each column, we get 2, 1, 2, 2, 2. Each person must be on duty in three columns. To have A and B on duty more than the others, the first column must be C, D, and E on duty. Since D and E are already scheduled, the 1st and 4th columns must be A, B, and C on duty. The 3rd column is still not fully scheduled, and B must work with everyone, so B is scheduled here.

A's 2nd, 6th, and 10th on-duty dates in December are 4, 10, and 16, respectively, so the five-digit number is 41016.

Therefore, the answer is 41016.