Simple Linear Equations, But Introducing Floors!

Since $\left \lfloor \dfrac{2x+1}{3} \right \rfloor + \left \lfloor \dfrac{4x+5}{6} \right \rfloor$ is an integer, so $\dfrac{3x-1}{2}$ must also be an integer.

Let $x= \dfrac{2n+1}{3}$ for some integer $n$ . $\text{ .... (1)}$

We'll also use the fact that:

$\dfrac{2x+1}{3} + \dfrac12 = \dfrac{4x+5}{6} \text{ .... (2)}$

For every real number $\alpha$ , we have:

$\alpha - \lfloor \alpha \rfloor < \dfrac12 \Rightarrow \left( \alpha + \dfrac12 \right) - \left \lfloor \alpha + \dfrac12 \right \rfloor \geq \dfrac12$

[that is, the fractional part of $\alpha < 1/2$ if and only if the fractional part of $\alpha + 1/2 \geq 1/2$ ] so that

$\left \lfloor \alpha \right \rfloor > \alpha - \dfrac12 \Rightarrow \left \lfloor \alpha + \dfrac12 \right \rfloor \leq \alpha$

From this, using $\left \lfloor y \right \rfloor \leq y$ , we obtain:

$\left \lfloor \alpha \right \rfloor + \left \lfloor \alpha + \dfrac12 \right \rfloor \leq 2\alpha \text{ ....(3)}$

And using $\left \lfloor y \right \rfloor > y-1$ we obtain:

$\left \lfloor \alpha \right \rfloor + \left \lfloor \alpha + \dfrac12 \right \rfloor > 2\alpha - 1 \text{ ....(4)}$

both for every real number $\alpha$ .

Putting inequalities (3) and (4) together and using (2) gives, for $\alpha = \dfrac{2x+1}{3}$ :

$\dfrac{2(2x+1)}{3} - 1 < \left \lfloor \dfrac{2x+1}{3} \right \rfloor + \left \lfloor \dfrac{4x+5}{6} \right \rfloor \leq \dfrac{2(2x+1)}{3}$

Using the original equation, this is:

$\dfrac{4x-1}{3} < \dfrac{3x-1}{2} \leq \dfrac{4x+2}{3}$ .

From this we obtain $1 < x \leq 7$ . Using (1) the only possibilities are:

$x = \dfrac53, \dfrac73, \dfrac93, \ldots, \dfrac{21}{3}$

This is exactly the set of all solutions, and their sum being $\boxed{39}$ .