Find the digits $A, B, C$ if $300A+30B+3C=111B$ 
If $300A+30B+3C=111B$, where $A$,$B$ and $C$ can take values from 0 to 9 digits, then what are the values of $A$, $B$ and $C$?

By trial and error the solution is $A=1, B=4, C=8$. My problem is how can I solve this through equations.
 A: Divide out by 3, getting $100A+10B+C = 37B$. Subtract to obtain $100A + C = 27B$. Now note that $100A + C$ is a multiple of 9 between 0 and 243 whose tens digit is 0. Since multiples of 9 have sums of digits equal to a multiple of 9, it is easy to see that the only possibilities for $100A + C$ are 0, 9, 108, and 207. Of these, only 0 and 108 are multiples of 27, so $27B$ is either 0 or 108. Plugging back in gives two solutions: $A=1,B=4,C=8$ and $A=B=C=0$.
A: Modulo $100$, 
$$\begin{align}
30B+3C
&\equiv11B\\
3C
&\equiv81B\\
C
&\equiv27B
\end{align}$$
Since $C$ is a digit, you can run through values of $B$ and quickly rule out all possibilities but two. For instance $B$ cannot be $1$ because $27$ is not a digit. And for instance $B$ cannot be $9$ because $27\cdot9=243$ and $43$ is not a digit. The only possibility is that $B$ must be $4$ or $0$, and consequently $C$ is $8$ or $0$ respectively. Substitute these in and solve for $A$, to find that $A$ is $1$ or $0$ respectively.
There are two solutions:
$$300(1)+30(4)+3(8)=(111)(4)$$ $$300(0)+30(0)+3(0)=(111)(0)$$
A: First notice that
$$
300A \le 300A + 3C = 81B
$$
i.e. $100A \le 27B$.
Since $B \le 9$ this immediately gives $\boldsymbol{A \in \{0,1,2\}}$.
In the other direction,
$$
81 B = 300A + 3C \le 300A + 27
$$
i.e. $27B \le 100A + 9$.
So $\boldsymbol{27B}$ is a multiple of nine between $\boldsymbol{100A}$ and $\boldsymbol{100A + 9}$ (inclusive).
We do casework on the value of $A \in \{0,1,2\}$:


*

*If $A = 0$, the only multiple of $27$ between $0$ and $9$ is $0$, which leads to $\boxed{A = B = C = 0}$.

*If $A = 1$, the only multiple of $27$ between $100$ and $109$ is $108 = 27 \cdot 4$, which leads to $\boxed{A = 1, B = 4, C = 8}$.

*If $A = 2$, there are no multiples of $27$ between $200$ and $209$.
So we have listed all possible solutions, and this also agrees with the other answers.
