A number theory problem by Prayas Rautray

The function $f(x) \; = \; \sum_{j=0}^5 \big\lfloor 2^j x\big\rfloor \hspace{2cm} x \in \mathbb{R}$ is an increasing function of $x$ , and $f(196) = 12348$ . If $196 - \tfrac{1}{32} < x < 196$ then it is clear that $\big\lfloor 2^j x \big\rfloor \; = \; 2^j \times 196 - 1 \hspace{2cm} 0 \le j \le 5$ and hence that $f(x) \,= \, f(196) - 6 \,=\, 12342$ . Thus $f(x) \le 12342$ for $x < 196$ and $f(x) \ge 12348$ for $x \ge 196$ , so the number of solutions to the equation $f(x) = 12345$ is $\boxed{0}$ .