2018 AMC 10B #25

Problem:

Let $\lfloor x\rfloor$ denote the greatest integer less than or equal to $x$. How many real numbers $x$ satisfy the equation $x^2+\mathrm{10,000}\lfloor x\rfloor =\mathrm{10,000}x$?

$\mathbf{\text{(A) }}197\mathbf{\text{(B) }}198\mathbf{\text{(C) }}199\mathbf{\text{(D) }}200\mathbf{\text{(E) }}201$

Solution:

Dividing the equation by $10000$, we get $$\frac{x^2}{10000}+\lfloor x\rfloor =x$$

Note that $\{x\}+\lfloor x\rfloor =x$, so we know that $$\frac{x^2}{10000}=\{x\}$$ Hence, the magnitude of this term must be less than $1$. $$\left|\frac{x^2}{10000}\right|<1$$

$$⟹-100<x<100$$

If $x$ is in this range, then $\lfloor x\rfloor$ is an integer between $-99$ and $99$, inclusive. For every value of $\lfloor x\rfloor$ in this range, we have a unique solution for $x$ from the given equation, so we have a total of $99-(-99)+1=199$ real solutions $⟶\boxed{{\displaystyle \mathbf{\text{(C)}}}}$.