How to get the value of $a + b + c$? $(0 \leq a  < b < c) \in Z$,
$a + b + c + ab + ac + bc + abc = 1622$
$a + b + c = ?$
I've assumed that $a = 0$, by that we can get rid of this part $ab + ac + abc$
Now $bc + b + c = 1622$.
But I found that was useless and got stuck. 

Hint:
$$a+b+c+ab+ac+bc+abc=(1+a)(1+b)(1+c)-1$$
Therefore $(1+a)(1+b)(1+c)=?$
