[Example 3.3.4] Divide 1983 into the sum of 12 positive integers, so that the product of their factorials is minimized.
First, note that the number of ways to divide 1983 into the sum of 12 positive integers is finite (at most $C_{1982}^{11}$ ways). Therefore, the number of corresponding products of factorials is also finite, so the minimum value must exist.
Let $a_{1}, a_{2}, \cdots, a_{12}$ be positive integers satisfying
$$
a_{1}+a_{2}+\cdots+a_{12}=1983,
$$

Our goal is to find the minimum value of
$$
p=a_{1}!a_{2}!\cdots a_{12}!
$$

We will study the direction of adjustment.
First, let's conduct a simulation to observe the pattern in which the product of factorials decreases.
By $1+5=2+4=3+3$,

and
$$
\begin{array}{l}
1!5!=120, \\
2!4!=48, \\
3!3!=36 .
\end{array}
$$

From the above experiment, we can see that if $a+b=c+d$, and $|b-a|>1$,
$$

We will study $a_{l}! a_{k}!$ and $\left(a_{l}-1\right)!\left(a_{k}+1\right)!$
$$
\begin{aligned}
a_{l}!a_{k}! & =a_{l}\left(a_{l}-1\right)!a_{k}! \\
& >\left(a_{l}-1\right)!\left(a_{k}+1\right) a_{k}!\quad\left(\because a_{l}>a_{k}+1\right) \\
& =\left(a_{l}-1\right)!\left(a_{k}+1\right)!.
\end{aligned}
$$

Here, the difference between $a_{l}-1$ and $a_{k}+1$ is smaller than the difference between $a_{l}$ and $a_{k}$. From the above, we can see that replacing $a_{l}! a_{k}!$ with $\left(a_{l}-1\right)!\left(a_{k}+1\right)!$ will result in a smaller product.
Therefore, the difference between any two numbers in $a_{1}, a_{2}, \cdots, a_{12}$ should be at most 1.
Thus, we can make the following adjustment:
If there are two numbers in $a_{1}, a_{2}, \cdots, a_{12}$ whose difference is greater than 1, then decrease the larger number by 1 and increase the smaller number by 1 to reduce the difference between these two numbers. This adjustment will result in a smaller product of factorials.

Continue this adjustment until the 12 numbers $a_{1}, a_{2}, \cdots, a_{12}$ are either equal or differ by 1.
Since
$$
1983=12 \times 165+3,
$$

we can set
$$
\begin{array}{c}
a_{1}=a_{2}=\cdots=a_{9}=165, \\
a_{10}=a_{11}=a_{12}=166 .
\end{array}
$$

At this point, $a_{1}!a_{2}!\cdots a_{12}!$ is minimized, and the minimum value is $(165!)^{9} \times(166!)^{3}$.