(Sârbă)
A BEAUTIFUL EQUATION WITH FLOOR FUNCTION
$[\alpha]$ denotes the largest integer less than or equal to $\alpha$ (floor function). An equation that uses all the numbers from 1 to 9, considered by us "beautiful" is
$$\left [ \frac{2x+1}{3}\right ]+\left [ \frac{4x+5}{6} \right ]=\frac{7x-8}{9}.$$
ANSWER CiP
$$x \in \left \{ -\frac{19}{7},\;-\frac{10}{7}\right \}$$
Solution CiP
We use characterization $[\alpha] \leqslant \alpha <[\alpha]+1$, so
$$[\alpha]=k\; \Leftrightarrow \;k \in \mathbb{Z}\;and\; k\leqslant \alpha<k+1.$$
The left side of the equation is an integer, so we must have $\frac{7x-8}{9}=k,\;k\in\mathbb{Z}.$ Hence
$$x=\frac{9k+8}{7}.\tag{1}$$
Applying the equivalent to the definition inequalities $\alpha-1<k\leqslant \alpha$ , we have
$$\frac{2x+1}{3}-1<\left [ \frac{2x+1}{3} \right ] \leqslant \frac{2x+1}{3},\tag{2}$$
$$\frac{4x+5}{6}-1<\left [ \frac{4x+5}{6} \right ] \leqslant \frac{4x+5}{6}.\tag{3}$$
Adding these two relationships and considering the equation, we get
$$\frac{2x+1}{3}+\frac{4x+5}{6}-2<\frac{7x-8}{9}\leqslant \frac{2x+1}{3}+\frac{4x+5}{6}.$$
The first inequality is successively equivalent to
$$\frac{2x+1}{3}+\frac{4x+5}{6}-\frac{7x-8}{9}<2\;\Leftrightarrow$$
$$\Leftrightarrow\;\frac{10x+37}{18}<2\;\Leftrightarrow\;x<-\frac{1}{10}.$$
The second inequality is successively equivalent to
$$\frac{2x+1}{3}+\frac{4x+5}{6}-\frac{7x-8}{9}\geqslant 0\;\Leftrightarrow$$
$$\Leftrightarrow\;\frac{10x+37}{18}\geqslant 0\;\Leftrightarrow \;x\geqslant -\frac{37}{10}.$$
So we have $-\frac{37}{10}\leqslant x <-\frac{1}{10}$ and considering the form (1) of $x$:
$$-\frac{37}{10}\leqslant \frac{9k+8}{7}<-\frac{1}{10}.$$
Solving these inequalities with the unknown $k$, we find $-\frac{113}{30}\leqslant k<-\frac{29}{30}.$ But the number $k$ being integer, we can only have $k \in \{-3,-2,-1\}$, obtaining for $x$, with the formula (1), respectively
$$x \in \left \{-\frac{19}{7}, -\frac{10}{7},-\frac{1}{7} \right \}.$$
But the value $x=-\frac{1}{7}$ does NOT check the equation (it appeared as a "foreign solution", summing two relations (2) and (3), but which are not independent of each other). We get the answer.
$\blacksquare$
REMARK CiP
$1^{R}$. Noting that for $\alpha =\frac{2x+1}{3}$ we have $\alpha +\frac{1}{2}=\frac{4x+5}{6}$, the left side of the given equation has the form $[\alpha ]+\left [\alpha +\frac{1}{2} \right ]$. But we have an identity of Hermite, we can say as beautiful
$$[\alpha]+\left [ \alpha +\frac{1}{2} \right ]=[2\cdot \alpha].$$
Then our equation is written equivalently
$$\left [ 2\cdot \frac{2x+1}{3}\right ]=\frac{7x-8}{9}\;\Leftrightarrow\;\left [ \frac{4x+2}{3} \right ]=\frac{7x-8}{9}$$
which is a more common type in problem collections.
Like before $x=\frac{9k+8}{7}$ and now it is enough to set the condition $k\leqslant \frac{4x+2}{3}<k+1\;\Leftrightarrow\;3k-2\leqslant 4x<3k+1\;\Leftrightarrow\;3k-2\leqslant \frac{36k+32}{7}<3k+1\;\Leftrightarrow$
$\Leftrightarrow\;-\frac{46}{15}\leqslant k<-\frac{5}{3}$, hence $k=-3$ or $k=-2$. We get the answer. The foreign solution stopped appearing.
(End Rem 1.)
$2^{R}$. Our problem is suggested by Problem 2088 [1995: 307] from the journal CRUX_MATHEMATICORUM. (The notation in right brackets means the year 1995, page 307. More specifically, it is volume 21, number 9, from November 1995, under the heading "PROBLEMS"). I modified, for aesthetic reasons, only the term on the right side of the equation. The solution appeared under the heading SOLUTIONS, in their coding [1996: 334].
(End Rem 2.)
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