Number theory - Find the number satisfying the condition. Find a 3 digit number which equals to 4 times the product of its digits. 

My approach:
I considered the Number to be $\overline{ABC}$ then wrote the relation $$100A+10B+C=4A\times B\times C$$ But i don't know how to proceed further. Please help.
 A: Hint: One good way to begin with is always to look at congruence class. For example it's easy to see that $4$ divides the RHS, hence it has to divide the LHS and in particular $4$ divides $c + 10b$. Thus $c$ is even which implies even a better condition: $8$ divides $100a + 10b + c$.
Another intresting condition is that if any digits were $5$ then your number is divisible by $10$ and so $c = 0$, but the RHS then implies that your number is $0$.
[from Barry Cipra observation]
Checking the equation mod $3$ we get
$$
  A + B + C = ABC \mod 3
$$
Assuming that $A,B,C$ are all non divisible by $3$, then we are left with 4 cases to check, but modulus $3$ they are all not possibile
$$
  \begin{array}{ll}
  1 + 1 + 1 = 0 \neq 1 = 1 \cdot 1 \cdot 1 
  & \quad 1 + 1 + 2 = 1 \neq 2 = 1 \cdot 1 \cdot 2 \\
  1 + 2 + 2 = 2 \neq 1 = 1 \cdot 2 \cdot 2
  & \quad 2 + 2 + 2 = 0 \neq 2 = 2 \cdot 2 \cdot 2
  \end{array}
$$
I encourage you to find more conditions, until eventually you are left with a feasible number of cases that can be checked by hand
A: Long solution without programming.
$100A+10B+C=4\times A\times B\times C (1)$
Divide both parts of (1) by 4
$25A+\frac{5}{2}B+\frac{C}{4}=A\times B\times C (2)$
From (2)
Based on Daniel Fischer suggestion
$C=2$ (2) goes to $25A+\frac{5}{2}B+\frac{2}{4}=2\times A\times B => 
\frac{50A+5B+1}{4}=A\times B$ and $50A+5B+1$ can not be divided by 4
$C=6$ (2) goes to $25A+\frac{5}{2}B+\frac{3}{2}=6\times A\times B => 
\frac{50A+5B+3}{12}=A\times B$ and $50A+5B+3$ can not be divided by 2
$C$ must be 4 or 8 and $B$ must be even.
If $c=8$ (2) can be rewritten as $25A+\frac{5}{2}B+2=8 \times A\times B (3')$
or
$\frac{25}{8}A+\frac{5}{16}B+\frac{1}{4}=A\times B (3'')$
(3'') can not be solved for for integers A and B because of sum plus $\frac{1}{4}$ and B must be divisible by 16 (wrong).
Try $C=4$ 
$25A+\frac{5}{2}B+1=4 \times A\times B (4')$ or
$\frac{25}{4}A+\frac{5}{8}B+\frac{1}{4}=A\times B (4'')$
Try (4'') with $B=2,4,6,8,0$
$B=0$ (4'') impossible
$B=2$ (4'') goes to $\frac{25}{4}A+\frac{5}{4}+\frac{1}{4}=2\times A (4''')$ or 
$25A+6=8A => 17A= -6$- impossible
$B=4$ (4'') goes to $\frac{25}{4}A+\frac{10}{4}+\frac{1}{4}=4\times A (4''')$ or 
$25A+11=16A => 9A= -11$- impossible
$B=6$ (4'') goes to $\frac{25}{4}A+\frac{5}{8}6+\frac{1}{4}=6\times A (4''')$ or 
$25A+16=24A => A= -16$- impossible
$B=8$ (4'') goes to $\frac{25}{4}A+\frac{5}{8}8+\frac{1}{4}=8\times A (4''')$ or 
$25A+21=32A => 7A= 21 => A=3$ - possible
384
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