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Relevant wiki: Floor and Ceiling Functions - Problem Solving

$\lfloor x \rfloor + |x| + \lceil x \rceil = 1000$

For this equation, we will analyze $3$ cases:

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Case 1 : $x$ is a non-integer real value

Note that for non-integer real values $x$ , we have

Integer $+$ Non-integer $+$ Integer $=$ Non-integer

We will never get an integer value result. This means that no non-integer real value $x$ satisfies this equation.

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Case 2 : $x$ is a non-negative integer

For non-negative integers $x$ , we have $\lfloor x \rfloor = \lceil x \rceil = |x| = x$ . This means we have

$x+x+x=1000\implies 3x=1000 \implies x=\dfrac{1000}{3}$

The value of $x$ found isn't an integer. Therefore, no non-negative integer $x$ satisfies this equation

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Case 3 : $x$ is a negative integer.

For negative integers $x$ , we have $\lfloor x \rfloor = \lceil x \rceil = x$ and $|x|=-x$ . This means we have

$x-x+x=1000\implies x=1000$

The value of $x$ found is non-negative. This means that no negative integer $x$ satisfies this equation either.

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From here, we can conclude that there are $\boxed{0}$ real values $x$ that satisfy this equation

Since $\lfloor x \rfloor, \lceil x \rceil$ and $1000$ are all intergers, $1000 - \lfloor x \rfloor - \lceil x \rceil = |x|$ is an integer too by closure and so is $x$ . Therefore $\lfloor x \rfloor = \lceil x \rceil = x$ . This simplifies the equation to $|x| + 2x = 1000$ . If $x$ is positive then this becomes $3x = 1000$ which has no integer solutions. If $x$ is negative then it becomes $x = 1000$ which is not a negative solution and thus rejected. If $x = 0$ then $0 = 1000$ which is obviously false. Therefore there are no solutions.

The only answer is $x=33\frac 1 3$ . The cieling and floor functions give 34 and 33. While these two and the sum are integers the absolute value function is $33\frac 1 3$ .
So the sum is not 1000 but $1000\frac 1 3$ . We also know integer+integer+non integer can not add up to an integer. So no solution.