2021 April MIMC 10 Problems/Problem 21

How many solutions are there for the equation $\left \lfloor{x}\right \rfloor^{2}-\left \lceil{x}\right \rceil=0$ . (Recall that $\left \lfloor{x}\right \rfloor$ is the largest integer less than $x$ , and $\left \lceil{x}\right \rceil$ is the smallest integer larger than $x$ .)

$\textbf{(A)} ~0 \qquad\textbf{(B)} ~1 \qquad\textbf{(C)} ~2 \qquad\textbf{(D)} ~3 \qquad\textbf{(E)} ~4 \qquad$

Solution

Let's split this question into different parts.

First case: $\left \lfloor{x}\right \rfloor=\left \lceil{x}\right \rceil$ . In this case, both $\left \lfloor{x}\right \rfloor$ and $\left \lceil{x}\right \rceil$ has to be equal to $x$ . Therefore, $\left \lfloor{x}\right \rfloor^{2}-\left \lceil{x}\right \rceil=x^2-x=0$ . Factoring this equation, we get that $x(x-1)=0$ which provides two solutions, $x=0$ and $x=1$ .

Second case: $\left \lfloor{x}\right \rfloor\neq\left \lceil{x}\right \rceil$ . In this case, $\left \lfloor{x}\right \rfloor$ must be equal to $\left \lceil{x}\right \rceil-1$ . Setting $y=\left \lfloor{x}\right \rfloor$ , we can get that $\left \lfloor{x}\right \rfloor^{2}-\left \lceil{x}\right \rceil=y^2-(y+1)=y^2-y-1=0$ . Calculating the discriminant which is $1^2-4\cdot1\cdot(-1)=5$ , we know that $y$ is an irrational number. However, since $y$ is a floor function of a number, it is always an integer. Therefore, there is no solution for this case.

In total, there are $\fbox{\textbf{(C)} 2}$ solutions for the equation.

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2021 April MIMC 10 Problems/Problem 21

Solution