A digit sum problem

When the number $x$ is doubled to obtain $2x$ , consider a digit $d$ in $x$ . If $0 \le d \le 4$ , then $0 \le 2d < 10$ , and so the digit $2d$ contributes $2d$ to $S(2x)$ . If $5 \le d \le 9$ , the $2d$ is a $2$ -digit number, and so its contribution to $S(2x)$ will be $2d-10$ (the unit digit) plus $1$ (the digit $1$ carried into the next column). No digit will have an effect producing more than one carry digit. Thus, if $x = \overline{a_1a_2a_3\cdots a_N}$ , then $S(x) \; = \; \sum_{j=1}^N a_j \hspace{2cm} S(2x) \; = \; \sum_{j=1}^N f(a_j)$ where $f(d) \; = \; \left\{ \begin{array}{lll} 2d & \hspace{1cm} & 0 \le d \le 4 \\ 2d-9 & & 5 \le d \le 9 \end{array} \right.$ It is easy to check that $d \le 5f(d)$ for all $0 \le d \le 9$ , with equality obtained with $f(5) = 1$ . Thus $S(x) \; = \; \sum_{j=1}^N x_j \; \le \; 5\sum_{j=1}^N f(x_j) \; = \; 5S(2x)$ for all integers $x$ , and hence $\frac{S(x)}{S(2x)} \le 5$ for all integers $x$ . Since $\frac{S(5)}{S(10)} = 5$ , this upper bound of $\boxed{5}$ is achieved.