Problem:
Let $\lfloor x\rfloor$ denote the greatest integer less than or equal to $x$. How many real numbers $x$ satisfy the equation $x^2 + 10{,}000\lfloor x \rfloor = 10{,}000x$?
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$
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 $1$. $$\left | \frac{x^2}{10000}\right |<1$$
$$\implies -100 < x < 100$$
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 $\longrightarrow \boxed{\textbf{(C)}}$.
Comments
Post a Comment