Difference between revisions of "2021 April MIMC 10 Problems/Problem 21"

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<math>\textbf{(A)} ~0 \qquad\textbf{(B)} ~1 \qquad\textbf{(C)} ~2 \qquad\textbf{(D)} ~3 \qquad\textbf{(E)} ~4 \qquad</math>
 
<math>\textbf{(A)} ~0 \qquad\textbf{(B)} ~1 \qquad\textbf{(C)} ~2 \qquad\textbf{(D)} ~3 \qquad\textbf{(E)} ~4 \qquad</math>
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==Solution==
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Let's split this question into different parts.
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First case: <math>\left \lfloor{x}\right \rfloor=\left \lceil{x}\right \rceil</math>. In this case, both <math>\left \lfloor{x}\right \rfloor</math> and <math>\left \lceil{x}\right \rceil</math> has to be equal to <math>x</math>. Therefore, <math>\left \lfloor{x}\right \rfloor^{2}-\left \lceil{x}\right \rceil=x^2-x=0</math>. Factoring this equation, we get that <math>x(x-1)=0</math> which provides two solutions, <math>x=0</math> and <math>x=1</math>.
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Second case: <math>\left \lfloor{x}\right \rfloor\neq\left \lceil{x}\right \rceil</math>. In this case, <math>\left \lfloor{x}\right \rfloor</math> must be equal to <math>\left \lceil{x}\right \rceil-1</math>. Setting <math>y=\left \lfloor{x}\right \rfloor</math>, we can get that <math>\left \lfloor{x}\right \rfloor^{2}-\left \lceil{x}\right \rceil=y^2-(y+1)=y^2-y-1=0</math>. Calculating the discriminant which is <math>1^2-4\cdot1\cdot(-1)=5</math>, we know that <math>y</math> is an irrational number. However, since <math>y</math> is a floor function of a number, it is always an integer. Therefore, there is no solution for this case.
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In total, there are <math>\fbox{\textbf{(C)} 2}</math> solutions for the equation.

Latest revision as of 14:00, 26 April 2021

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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