<s>[INST] The number of all 3-digit numbers abc for which abc + ab+bc + ac+a+b+c = 29 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.
 [/INST] The prime fctorisation of $30$ is $5\times3\times2$. The only ordered 3-uples for $a+1,b+1$ and $c+1$  where $a+1 \in [2,10]$ and $b+1,c+1 \in [1,10]$ are : $$(2,3,5),(2,5,3),(3,1,10),(3,2,5),(3,5,2),(3,10,1),(5,1,6),(5,2,3),(5,3,2),(5,6,1),(6,1,5),(6,5,1),(10,1,3),(10,3,1)$$
So the solutions are : $$124,142,209,214,241,290,405,412,421,450,504,540,902,920$$
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