Problem:
Define $\lfloor x \rfloor$ as the greatest integer less than or equal to $x$ and $\{x\}=x- \lfloor x \rfloor$. How many real solutions are there to $$\frac{\{x\}}{\lfloor x \rfloor} = \frac{x}{100}?$$
Solution:
The given condition can be rewritten as $$\frac{\{x\}}{\lfloor x \rfloor} = \frac{\{x\} + \lfloor x \rfloor}{100}$$
Manipulating, we get $$\{x\} = \frac{\lfloor x \rfloor^2}{\lfloor x \rfloor - 100} \qquad (1)$$
Since $\{x\}$ is the decimal part of $x$, it must be less than $1$. That is, we must have $$\frac{\lfloor x \rfloor^2}{\lfloor x \rfloor - 100} < 1$$
$$\implies \frac{-\sqrt{401} - 1}{2} < \lfloor x \rfloor < \frac{\sqrt{401} - 1}{2}$$
$$\implies -10.5 < \lfloor x \rfloor < 9.5 \qquad (2)$$
For each value of $\lfloor x \rfloor$ in this range, we have exactly one possible value of $\{x\}$, which can be found from $(1)$.
Lastly, $\lfloor x \rfloor$ must take on an integer value in the range $(2)$, so the following values work: $$-10, -9, \dots, -1, 1, 2, \dots, 9$$
These are a total of $10+9=\boxed{19}$ values, which is our answer. $\square$
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