Find the smallest 3 digit number which satisfies the following conditions 
The smallest 3 digit number n such that if the 3 digits are $a,b$ and $c$, then 
$n = a+b+c+ab+bc+ac+abc$.

I tried $n = 100a + 10b + c$ (say)
Plugged that into the given equation but couldn't simplify it further. I have no idea how should I proceed here.
Any help would be appreciated.
 A: To find a lower bound for the answer we calculate necessary conditions for $a = 1$ first. Then we want to solve $$100 + 10b + c = 1 + b + c + b + bc + c+ bc = 1+ 2b + 2c+ 2bc,$$
i.e. $99 + 8b =c + 2bc = (1+2b)c $ and thus $$c = \frac{99+8b}{1+2b}.$$
From this we obtain $1+2b \mid 99 + 8b$, so $1+2b \mid 99+8b - 4\cdot (1+2b) = 95 = 5 \cdot 19$. Since $b \in \{0, \dots, 9\}$ this can only happen if $1+2b \in \{1,5, 19\}$, i.e. $b \in \{0, 2,9\}$. Now we see that $b \in \{0,2\}$ yields $c \in \{99, 23\}$, so those two possibilities do not work. For $b = 9$ we get $$c = \frac{99 + 72}{1 + 18} = \frac{171}{19} = 9$$ and hence $199$ is the smallest possible solution with $a>0$. As pointed out in the comments by lulu, this actually is a solution.
A: Making use of lulu's astute observation in comments that $n=a99$ works for any $a$, it remains to show that $199$ is the only solution with $a=1$. To do this note that with $a=1$, the condition becomes $100+10b+c=1+b+c+b+bc+c+bc$, or
$$99+8b=c(2b+1)$$
which implies $c$ is odd. Writing $c=2d-1$ with $1\le d\le5$ and doing a little algebra, the equation now becomes
$$5(b+10)=d(2b+1)$$
and this implies either $d=5$ or $b=2$ or $b=7$ (i.e., either $5\mid d$ or $5\mid(2b+1)$).  The possibilities $b=2$ and $b=7$ are easily ruled out, leaving only $d=5$, which gives $c=9$ and $b=9$ (from $b+10=2b+1$).
