Question Number 151101 by mathdanisur last updated on 18/Aug/21
![Solve for real numbers the equation [(x/2)] + [((3x)/5)] = [(x/(10))] + x , where we denoting by [x] the great integer part of x.](Q151101.png)
$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\left[\frac{\mathrm{x}}{\mathrm{2}}\right]\:+\:\left[\frac{\mathrm{3x}}{\mathrm{5}}\right]\:=\:\left[\frac{\mathrm{x}}{\mathrm{10}}\right]\:+\:\mathrm{x}\:,\:\:\mathrm{where}\:\mathrm{we} \\ $$$$\mathrm{denoting}\:\mathrm{by}\:\left[\boldsymbol{\mathrm{x}}\right]\:\mathrm{the}\:\mathrm{great}\:\mathrm{integer}\:\mathrm{part} \\ $$$$\mathrm{of}\:\boldsymbol{\mathrm{x}}. \\ $$
Answered by dumitrel last updated on 18/Aug/21
![⇒x∈Z⇒x=10c+r,c∈Z,r∈{0,1,...,9} ⇒[5c+(r/2)]+[6c+((3r)/5)]=[c+(r/(10))]+10c+r 5c+[(r/2)]+6c+[((3r)/5)]=c+[(r/(10))]+10c+r [(r/2)]+[((3r)/5)]=[(r/(10))]+r⇒r∉{1,3}⇒ x=10c+r,r∈{0,2,4,5,6,7,8,9}](Q151108.png)
$$\Rightarrow{x}\in{Z}\Rightarrow{x}=\mathrm{10}{c}+{r},{c}\in{Z},{r}\in\left\{\mathrm{0},\mathrm{1},...,\mathrm{9}\right\} \\ $$$$\Rightarrow\left[\mathrm{5}{c}+\frac{{r}}{\mathrm{2}}\right]+\left[\mathrm{6}{c}+\frac{\mathrm{3}{r}}{\mathrm{5}}\right]=\left[{c}+\frac{{r}}{\mathrm{10}}\right]+\mathrm{10}{c}+{r} \\ $$$$\mathrm{5}{c}+\left[\frac{{r}}{\mathrm{2}}\right]+\mathrm{6}{c}+\left[\frac{\mathrm{3}{r}}{\mathrm{5}}\right]={c}+\left[\frac{{r}}{\mathrm{10}}\right]+\mathrm{10}{c}+{r} \\ $$$$\left[\frac{{r}}{\mathrm{2}}\right]+\left[\frac{\mathrm{3}{r}}{\mathrm{5}}\right]=\left[\frac{{r}}{\mathrm{10}}\right]+{r}\Rightarrow{r}\notin\left\{\mathrm{1},\mathrm{3}\right\}\Rightarrow \\ $$$${x}=\mathrm{10}{c}+{r},{r}\in\left\{\mathrm{0},\mathrm{2},\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{7},\mathrm{8},\mathrm{9}\right\} \\ $$$$ \\ $$
Commented by mathdanisur last updated on 18/Aug/21
$$\mathrm{nice}\:\boldsymbol{\mathrm{S}}\mathrm{er}\:\mathrm{thank}\:\mathrm{you} \\ $$