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Question Number 86009 by M±th+et£s last updated on 26/Mar/20

$$solveinR:\left[\frac{x}{2}\right]+\left[\frac{2x}{3}\right]-x=0$$

Answered by MJS last updated on 26/Mar/20

$$\left[\frac{x}{2}\right]\in \mathbb{Z}\wedge \left[\frac{2x}{3}\right]\in \mathbb{Z}\Rightarrow x\in \mathbb{Z}$$$$\mathrm{let}k,m,n\in \mathbb{Z}$$$$\left[\frac{x}{2}\right]=\{\begin{array}{l}\frac{x}{2};x=2m\\ \frac{x}{2}-\frac{1}{2};x=2m+1\end{array}$$$$\left[\frac{2x}{3}\right]=\{\begin{array}{l}\frac{2x}{3};x=3n\\ \frac{2x}{3}-\frac{1}{3};x=3n+2\\ \frac{2x}{3}-\frac{2}{3};2x=3n+1\end{array}$$$$\left(1\right)x=2m=3n\Rightarrow x=6k$$$$[\frac{6k}{2}[+\left[\frac{12k}{3}\right]=7k$$$$7k=6k\Rightarrow k=0\Rightarrow x=0\bullet$$$$\left(2\right)x=2m=3n+2\Rightarrow x=6k+2$$$$\left[\frac{6k+2}{2}\right]+\left[\frac{12k+4}{3}\right]=7k+2$$$$7k+2=6k+2\Rightarrow k=0\Rightarrow x=2\bullet$$$$\left(3\right)x=2m=3n+1\Rightarrow x=6k+4$$$$\left[\frac{6k+4}{2}\right]+\left[\frac{12k+8}{3}\right]=7k+4$$$$7k+4=6k+4\Rightarrow k=0\Rightarrow x=4\bullet$$$$\left(4\right)x=2m+1=3n\Rightarrow x=6k+3$$$$\left[\frac{6k+3}{2}\right]+\left[\frac{12k+6}{3}\right]=7k+3$$$$7k+3=6k+3\Rightarrow k=0\Rightarrow x=3\bullet$$$$\left(5\right)x=2m+1=3n+2\Rightarrow x=6k+5$$$$\left[\frac{6k+5}{2}\right]+\left[\frac{12k+10}{3}\right]=7k+5$$$$7k+5=6k+5\Rightarrow k=0\Rightarrow x=5\bullet$$$$\left(6\right)x=2m+1=3n+1\Rightarrow x=6k+1$$$$\left[\frac{6k+1}{2}\right]+\left[\frac{12k+2}{3}\right]=7k$$$$7k=6k+1\Rightarrow k=1\Rightarrow x=7\bullet$$$$$$$$\mathrm{answer}\mathrm{is}x\in \{0,2,3,4,5,7\}$$

Commented by M±th+et£s last updated on 26/Mar/20

$$thankyousirgodblessyou$$

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