10.86 Try to divide the seven digits $3,4,5,6,7,8,9$ into two groups, and arrange them into a three-digit number and a four-digit number, respectively, such that the product of these two numbers is maximized. How should the groups be divided and arranged, and prove your conclusion.
(China Beijing Junior High School Grade 3 Mathematics Competition, 1981)
[Solution] When forming each number, the larger digit should be placed at the highest position. Thus, 9 and 8 should be the leading digits of the two numbers, respectively.

Now, select one digit from the remaining digits to add to 9 and 8 to form two two-digit numbers, and make their product as large as possible. Clearly, 7 and 6 should be chosen.

Since $96 \cdot 87 > 97 \cdot 86$, among the two-digit numbers formed by two digits, the product of 96 and 87 is the largest.
From this, we consider the general case:
Let $A$ and $B$ be natural numbers, with $A > B$. Now, add the digits $c$ and $d$ ($c > d$) to $A$ and $B$ to form new numbers. There are only two possible groups:
First group: $\quad \overline{A_{c}} = 10A + c, \quad \overline{B d} = 10B + d$.
Second group: $\quad \overline{A d} = 10A + d, \quad \overline{B c} = 10B + c$.
Now, compare the products of these two groups of new numbers:
$$
\begin{aligned}
& \overline{A c} \cdot \overline{B d} - \overline{A d} \cdot \overline{B c} \\
= & (10A + c)(10B + d) - (10A + d)(10B + c) \\
= & 10(A d + B c - A c - B d) \\
= & 10(A - B)(d - c) < 0 .
\end{aligned}
$$

Therefore,
$$
\overline{A c} \cdot \overline{B d} < \overline{A d} \cdot \overline{B c}.
$$

From this, we know that when dividing the four digits 9, 8, 7, and 6 into two groups to form two two-digit numbers, the first group should take 9, then the second group should take 8, and the first group should take the smaller of 7 and 6, which is 6, and the second group should take 7. This way, the product of the two two-digit numbers $96 \cdot 87$ is the largest.

Similarly, for the six digits 9, 8, 7, 6, 5, and 4, divide them into two groups to form two three-digit numbers. After the first group takes 96 and the second group takes 87, the first group should take the smaller of 5 and 4, which is 4, and the second group should take 5. This way, the product of the two three-digit numbers $964 \cdot 875$ is the largest.

Finally, there is one digit 3 left. We can add another digit 0, and add 3 and 0 to 964 and 875, respectively. According to the above conclusion, the two numbers with the largest product should be $9640 \cdot 8753$. Removing the last digit 0, forming a three-digit number and a four-digit number, their product is the largest, which is 964 and 8753.

Therefore, dividing the seven digits 3, 4, 5, 6, 7, 8, and 9 into two groups to form a three-digit number 964 and a four-digit number 8753, the product $964 \cdot 8753$ is the largest.