10.6 If the same letter represents the same digit, try to decipher the number puzzle:
$$
\overline{A A} \cdot \overline{A B}=\overline{C D E F}, \quad \overline{C C} \cdot \overline{C E F}=\overline{C D E F} .
$$
(Kyiv Mathematical Olympiad, 1980)
[Solution] Since $\overline{C E F} \geqslant C \cdot 10^{2}, \overline{C D E F}\overline{C C} \cdot \overline{C E F} \geqslant \overline{C C} \cdot C \cdot 10^{2}=C^{2} \cdot 11 \cdot 10^{2} . \\
(C+1) \cdot 10>11 C^{2} .
\end{array}
$

The solution satisfying this inequality is
$
\frac{5-3 \sqrt{15}}{11}\overline{1 D E 0} .
$

Therefore,
$
A \neq 5 \text {. }
$

If $B=5, A$ is even, then we have
$
\begin{array}{c}
11 \cdot A \cdot \overline{A 5}=\overline{A A} \cdot \overline{A 5}=\overline{1 D E 0}=11 \cdot \overline{1 E F}, \\
A \cdot \overline{A 5}=\overline{1 E F}=\overline{1 E 0} .
\end{array}
$

Since when $A \leqslant 2$, $A \cdot \overline{A 5}200$.
Thus,
$
2<A<5 \text {. }
$

Then the even number
$
A=4 \text {. }
$

Therefore, the first equation is
$
44 \cdot 45=\overline{1 D E 0}=1980 \text {. }
$

Thus,
$
D=9, \quad E=8 \text {. }
$

So one solution to this number puzzle is
$
\begin{array}{l}
44 \cdot 45=1980, \\
11 \cdot 180=1980 .
\end{array}
$

Clearly, $A \neq 0$. If $B=0$, then
$
\begin{aligned}
11 \cdot A \cdot \overline{A 0} & =\overline{1 D E 0}=11 \cdot \overline{1 E 0}, \\
A^{2} & =\overline{1 E},
\end{aligned}
$

Thus, $A=4, \quad E=6, \quad D=7$.
In this case, another solution to this number puzzle is
$
\begin{array}{l}
44 \cdot 40=1760, \\
11 \cdot 160=1760 .
\end{array}
$

In summary, this number puzzle has two solutions.