10.8 In a game, the "magician" asks a person to think of a three-digit number $\overline{a b c}$ (where $a, b, c$ are the decimal digits of the number in order), and asks this person to select 5 numbers $\overline{a c b}, \overline{b a c}, \overline{b c a}, \overline{c a b}$ and $\overline{b a}$, and find the sum $N$ of these 5 numbers, and tell the sum $N$ to the "magician". Then, the "magician" can reveal the number (abc) that the person thought of. Now, let $N=3194$. Please act as the "magician" and find the number $\overline{a b c}$.
(4th American Invitational Mathematics Examination, 1986)
［Solution］ Since
$$
\begin{aligned}
& \overline{a b c}+N \\
= & \overline{a b c}+\overline{a c b}+\overline{b a c}+\overline{b c a}+\overline{c a b}+\overline{c b a} \\
= & 222(a+b+c) .
\end{aligned}
$$

We should find a multiple of 222, $222k$, such that it is greater than $N$ (since $\overline{a b c} \neq 0$), and less than $N+1000$ (since $\overline{a b c}$ is a three-digit number), and the sum of the digits of $222k-N$ is $k$. In this way, $222k-N$ is the solution $(a b c)$.
That is,
$$
N<222k<N+1000 \text{. }
$$

Since $N=3194$, and
$$
14 \cdot 222<3194<3194+1000<19 \cdot 222 \text{. }
$$

Thus, the possible values of $k$ are $15, 16, 17, 18$.
We will check each one. Since
$222 \cdot 15-3194=136$ has a sum of digits of $10 \neq 15$,
$222 \cdot 16-3194=358$ has a sum of digits of $16=16$,
$222 \cdot 17-3194=580$ has a sum of digits of $13 \neq 17$,
$222 \cdot 18-3194=802$ has a sum of digits of $10 \neq 18$.
Thus,
$$
k=16 \text{. }
$$

The solution $(a b c)$ is
$$
222 \cdot 16-3194=358 \text{. }
$$