Linear Sum

Note that for $x<0$ the series diverges, so for the problem, we will assume $x \geq 0.$ Then $\begin{aligned} y &= f(x) \\ &= \sum_{n=1}^\infty \left\lfloor\frac{x}{5^n}\right\rfloor \\ &= \sum_{n=1}^{\lfloor \log_5x\rfloor} \frac{x}{5^n} - \left\{\frac{x}{5^n}\right\} \\ &= \frac{x}{4} - \frac{x}{4}\cdot 5^{-\lfloor\log_5 x\rfloor} - \sum_{n=1}^{\lfloor\log_5 x\rfloor} \left\{\frac{x}{5^n}\right\} \end{aligned}$

Zooming out the graph is a uniform scaling of both axes, which means we should consider substitute $(rx, ry)$ for $(x,y)$ and let $r\to\infty$ . Doing this gives $ry = \frac{rx}{4} - \frac{rx}{4}\cdot 5^{-\lfloor\log_5 rx\rfloor} - \sum_{n=1}^{\lfloor\log_5 rx\rfloor} \left\{\frac{rx}{5^n}\right\}$ Dividing by $r$ and subtracting $\frac{x}{4}$ from both sides before applying the absolute value: $\begin{aligned} \left|y-\frac{x}{4}\right| &= \left|\frac{x}{4}\cdot 5^{-\lfloor\log_5 rx\rfloor} + \frac{1}{r}\sum_{n=1}^{\lfloor\log_5 rx\rfloor} \left\{\frac{rx}{5^n}\right\}\right| \\ &\leq \frac{x}{4}\cdot 5^{-\lfloor\log_5 rx\rfloor} + \frac{\lfloor\log_5 rx\rfloor}{r} \xrightarrow{r\to\infty} 0, \qquad \text{ for any fixed } x \end{aligned}$

It follows that the apparent graph by scaling is $y = \frac{x}{4}, \qquad \implies \qquad m=4, b=0$ so that the answer is $m - b = \boxed{4}$

---

Remark: Note that this is really the only reasonable interpretation, where we scale all our variables by a scale factor $r$ and then redefine our unit along the axes to be that same scale factor $r$ . Otherwise, we would have to deal with the fact that all values of $b$ actually look like zero after enough scaling, which sounds like a good thing but it actually means that the value of $b$ in the problem can be considered to be undefined unless we redefine the unit