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So, every question in here regarding these kinds of equations only involves natural numbers. Generally, I'm looking for a method to solve equations on the form: $\lfloor ax+b\rfloor=c$ for $a,b\in\mathbb{Q},\,c\in\mathbb{Z}$. If this proves too troublesome, a method for approximating a solution or an interval containing the solution would be adequate.

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  • $\begingroup$ if $\lfloor ax + b \rfloor = c$ then by definition $c \in \mathbb Z$. $\endgroup$ – fleablood Jun 14 at 14:52
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Well that'd just be a mater of solving $c \le ax + b < c + 1$ so...

$c-b \le ax < c + 1 - b$ and if $a > 0$ then

$\frac {c-b}{a}\le x < \frac {c+1-b}a$ and if $a < 0$ then

$\frac {c-b}{a}\ge x > \frac {c+1-b}a$

That's all there is to it.

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  • $\begingroup$ As a follow-up question, given an equation, say $\lfloor \alpha\lfloor ax+b\rfloor+\beta\rfloor=c$ for $a,b,\alpha,\beta\in\mathbb{Q},\, c\in\mathbb{Z}$, would it still be possible to find an interval of solutions in general? $\endgroup$ – Kristian S. Jensen Jun 14 at 16:21
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First of all, since $\lfloor ax+b\rfloor=c$, $c$ is not only $\in \Bbb{Q}$, but $c \in \Bbb{Z}$. Secondly, $$\lfloor ax+b\rfloor=c \leftrightarrow c \le ax+b < c+1 $$ Which translates to two inequatlities: $$ax+b \ge c, \text{ and} \\ ax+b<c+1$$ Any $x \in \Bbb{R}$ that satisfies the above two inequalities will be your solution to $\lfloor ax+b\rfloor=c.$

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