Number equal to the sum of digits + product of digits) Are every number equal to (sum of digits + product of digits) in a given base only two digits long ? 
Thought about limiting like this :
$$b^{(n - 1)} \leq N = \text{Product digits} + \text{Sum digits} \leq (b - 1)*n + (b - 1)^{(n)}$$
Obviously the set is finite (when $n \rightarrow +\infty$) but can't figure out how to get more precision about the set
Any idea ?
ie in base 10 : {0,19,29,39,49....,99}
Edit : 'base' abreviation : b. Number of digits for a number N : n 

Bear with me:
For $k \ge 1; b\ge 2$ $(b - 1)^k \le b^k + (-1)^k$. Pf: induction: For $k =1$ then $(b-1)^1 = b^1 + (-1)^1$. For $k = 2$ then $(b-1)^2 = b^2 -2b +1 < b^2 + 1$. If $(b -1)^n \le b^n + (-1)^n$ then $$(b-1)^{n+1} = (b-1)^n(b-1) \le (b^n + (-1)^n)(b-1) = b^{n+1} + (-1)^nb - b^n +(-1)^{n+1} = b^{n+1} + (-1)^{n+1} - b(b^{n-1} + (-1)^n) \ge b^{n+1} + (-1)^{n+1}$$. With equality holding only if $k = 1$.  But for $k = 0 ; (b-1)^0 = 1 > 0 = b^0 - 1^0$.
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Let $n = \sum_{i=0}^k a_i b^k$ be a $k+1$ digit number and $S = \sum_{i=0}^k a_i$
$n - S = \sum_{i=0}^k a_i (b^k - 1) \ge \sum_{i=1}^k a_i (b - 1)^k$.
(we must can/must index from $1$ on the RHS because $a_0*(b^k -1) = 0$)
Now each $a_i \le (b-1)$ so $P = \prod_{i=0}^k a_i = a_k*(b-1)^k$
So $n - S \ge \sum_{i=1}^{k}a_i(b-1)^k$ 
$= a_k*(b-1)^k+ \sum_{j=1}^{k-1}a_j(b-1)^k$
$\ge P$
And equality only holds if
i) $\sum_{j=1}^{k-1}a_j(b-1)^k= 0$
either a) all $a_j; 1\le j \le k$ are equal to $0$ or b) $k -1 < 1$ i.e. $k < 2$ and $n $ is a one or two digit number.
If a) then $P =0$ and $n - S=0$ which can only happen if $n$ is a one digit number. ($a_i*b^i > a_i$ if $a_i > 0; i > 0$).  So b) must hold.
b) If $n$ is a one- digit number then $n - S = 0 = P$ so $P = 0$ and $n= 0$.
So this only true if $n$ is a two digit number or $n = 0$.
ii) Furthermore, If $n$ is a two-digit number this is only true if:
$n = a_1*b + a_0 = a_1 + a_0 + a_1*a_0; a_1 \ne 0$
$a_1(b - (a_0-1)) = 0$
$a_1 \ne 0$  so $a_0 = b- 1$
So the complete set of solutions for $n$ are $\{0, a_1*b + (b-1)|1 \le a_1 < b\}$
