Summing of product of digits

• one dig numbers: $\\ p_1 = \sum\limits_{i=1}^9 p(i) = \sum\limits_{i=1}^9 i = 45$

• two digits numbers: $\\ p_2 = \sum\limits_{i=10}^{99} p(i) \\ = \sum\limits_{i=1}^9 p(10i) + \sum\limits_{i=1}^9 \sum\limits_{j=1}^9 p(10i+j) \\ = \sum\limits_{i=1}^9 i + \sum\limits_{i=1}^9 \sum\limits_{j=1}^9 (i \times j) \\ = \sum\limits_{i=1}^9 i + \sum\limits_{i=1}^9 i \times \sum\limits_{j=1}^9 j \\ = p_1 + p_1 \times p_1 \\ = p_1(p_1 + 1)$

• three digits numbers: $\\ p_3 = \sum\limits_{i=100}^{999} p(i) \\ = \sum\limits_{i=1}^9 (\sum\limits_{j=0}^{99} p(100i + j)) \\ = \sum\limits_{i=1}^9 (p(100i) + \sum\limits_{j=1}^{99} p(100i + j)) \\ = \sum\limits_{i=1}^9 p(100i) + \sum\limits_{i=1}^9 (\sum\limits_{j=1}^{99} p(100i + j)) \\ = p_1 + \sum\limits_{i=1}^9 (i \times \sum\limits_{j=1}^{99} p(j)) \\ = p_1 + \sum\limits_{i=1}^9 (i \times (p_1 + p_2)) \\ = p_1 + p_2 \times \sum\limits_{i=1}^9 i \\ = p_1 + (p_1 + p_2) \times p_1 \\ = p_1 \times (p_1 +1)^2$

So $\sum\limits_{i=1}^{999} p(i) = p_1 + p_2 + p_3 \\ = 45 + (45 \times 46) + 45 \times (45 + 1)^2 \\ = 45 \times (1 + 46 + 46^2) \\ = 45 \times 2163 = 45 \times 21 \times 103 \\ = 3^3 \times 5 \times 7 \times 103$

Note that $p(n)$ can be defined as product of digits of number $n$ with all $0$ digit replaced with $1$ . Let $S(n) = \displaystyle \sum_{k=1}^n p(k)$ . Then we have:

$\begin{aligned} S(999) & = \sum_{k=1}^{999} p(k) \\ & = \sum_{k=1}^9 p(k) + \sum_{k=10}^{99} p(k) + \sum_{k=100}^{999} p(k) \\ & = \sum_{k=1}^9 k + \sum_{k=1}^9 k(\red 1 + 1 + 2 + \cdots + 9) + \sum_{k=100}^{999} p(k) & \small \red{\text{0 replaced with 1}} \\ & = \blue{S(9) + S(9)(1+S(9))} + \sum_{k=1}^9 p(100k) + \sum_{k=1}^9 kS(99) & \small \blue{\text{Note that }S(99) = S(9)(S(9)+2)} \\ & = S(9)(S(9)+2) + S(9) + S(9)^2(S(9)+2) \\ & = S(9)(S(9)+2 + 1 + S(9)(S(9)+2)) & \small \blue{\text{and }S(9) = \sum_{k=1}^9 k = \frac {9(9+1)}2= 45} \\ & = 45 (2163) \\ & = 3^3\cdot 5 \cdot 7 \cdot 103 \end{aligned}$

Therefore the greatest prime divisor of $S(999)$ is $\boxed{103}$ .