$1 \cdot 187$ In a game, the "magician" asks a person to think of a three-digit number $(abc)$ (where $a, b, c$ are the decimal digits of the number in sequence), and asks this person to select 5 numbers $(acb), (bac), (bca), (cab)$, and $(cba)$, and find the sum $N$ of these 5 numbers, and tell the sum $N$ to the "magician". Then, the "magician" can say the number $(abc)$ that the person thought of.
Now let $N=3194$, please be the "magician" and find the number $(abc)$.

Given $N=3194$, let's solve for the number $(abc)$.
$$
\text { [Solution] } \begin{aligned}
& (a b c)+N \\
= & (a b c)+(a c b)+(b a c)+(b c a)+(c a b)+(c b a) \\
= & 222(a+b+c) .
\end{aligned}
$$

Therefore, we should look for a multiple of 222, $222k$, such that it is greater than $N$ (since $(a b c) \neq 0$), and less than $N+1000$ (since $(a b c)$ is a three-digit number), and the sum of the digits of $222k - N$ is $k$. In this way, $222k - N$ is the solution we are looking for $(a b c)$.
Since $N=3194$, and
$$
14 \times 222 < 3194 < 19 \times 222,
$$

$k$ could be $15, 16, 17, 18$.
Since $222 \times 15 - 3194 = 136$ has a sum of digits of 10, which is not equal to 15, $222 \times 16 - 3194 = 358$ has a sum of digits of 16,
$222 \times 17 - 3194 = 580$ has a sum of digits of 13, which is not equal to 17,
$222 \times 18 - 3194 = 802$ has a sum of digits of 10, which is not equal to 18, thus $k = 16$.
$$
\begin{aligned}
\text { The required }(a b c) & = 222 \times 16 - 3194 \\
& = 358 .
\end{aligned}
$$