The number of all 3-digit numbers abc (in base 10) for which $\ abc+ab+bc+ac+a+b+c = 29$ is
(A) 6
(B) 10
(C) 14
(D) 18
My working:
$\ ab (c + 1) +b (c + 1) + a(c + 1) + c + 1 = 30$
$\ (a + 1) (b +1) (c+1) = 30 $
$\ 9>a>0$
$\ 0\le b,c<9$
The problem I am facing:
I don't know how to figure out the no. of solutions for which this is true.