# 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> | ||

==Solution== | ==Solution== | ||

− | + | Let's split this question into different parts. | |

+ | |||

+ | 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>. | ||

+ | |||

+ | 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. | ||

+ | |||

+ | 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 . (Recall that is the largest integer less than , and is the smallest integer larger than .)

## Solution

Let's split this question into different parts.

First case: . In this case, both and has to be equal to . Therefore, . Factoring this equation, we get that which provides two solutions, and .

Second case: . In this case, must be equal to . Setting , we can get that . Calculating the discriminant which is , we know that is an irrational number. However, since is a floor function of a number, it is always an integer. Therefore, there is no solution for this case.

In total, there are solutions for the equation.