2021 DMC 10 #18

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 $$\{x\}=\frac{\lfloor x{\rfloor }^2}{\lfloor x\rfloor -100}(1)$$

Since $\{x\}$ is the decimal part of $x$, it must be less than $1$. That is, we must have $$\frac{\lfloor x{\rfloor }^2}{\lfloor x\rfloor -100}<1$$

$$⟹\frac{-\sqrt{401}-1}{2}<\lfloor x\rfloor <\frac{\sqrt{401}-1}{2}$$

$$⟹-10.5<\lfloor x\rfloor <9.5(2)$$

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 $10+9=\boxed{{\displaystyle 19}}$ values, which is our answer. $◻$