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Question Number 86009 by M±th+et£s last updated on 26/Mar/20
solve in R :[(x/2)]+[((2x)/3)]−x=0
$${solve}\:{in}\:{R}\::\left[\frac{{x}}{\mathrm{2}}\right]+\left[\frac{\mathrm{2}{x}}{\mathrm{3}}\right]−{x}=\mathrm{0} \\ $$
Answered by MJS last updated on 26/Mar/20
[(x/2)]∈Z∧[((2x)/3)]∈Z⇒x∈Z  let k, m, n ∈Z  [(x/2)]= { (((x/2); x=2m)),(((x/2)−(1/2); x=2m+1)) :}  [((2x)/3)]= { ((((2x)/3); x=3n)),((((2x)/3)−(1/3); x=3n+2)),((((2x)/3)−(2/3); 2x=3n+1)) :}  (1) x=2m=3n ⇒  x=6k       [((6k)/2)[+[((12k)/3)]=7k       7k=6k ⇒ k=0 ⇒ x=0 •  (2) x=2m=3n+2 ⇒ x=6k+2       [((6k+2)/2)]+[((12k+4)/3)]=7k+2       7k+2=6k+2 ⇒ k=0 ⇒ x=2 •  (3) x=2m=3n+1 ⇒ x=6k+4       [((6k+4)/2)]+[((12k+8)/3)]=7k+4       7k+4=6k+4 ⇒ k=0 ⇒ x=4 •  (4) x=2m+1=3n ⇒ x=6k+3       [((6k+3)/2)]+[((12k+6)/3)]=7k+3       7k+3=6k+3 ⇒ k=0 ⇒ x=3 •  (5) x=2m+1=3n+2 ⇒ x=6k+5       [((6k+5)/2)]+[((12k+10)/3)]=7k+5       7k+5=6k+5 ⇒ k=0 ⇒ x=5 •  (6) x=2m+1=3n+1 ⇒ x=6k+1       [((6k+1)/2)]+[((12k+2)/3)]=7k       7k=6k+1 ⇒ k=1 ⇒ x=7 •    answer is x∈{0, 2, 3, 4, 5, 7}
$$\left[\frac{{x}}{\mathrm{2}}\right]\in\mathbb{Z}\wedge\left[\frac{\mathrm{2}{x}}{\mathrm{3}}\right]\in\mathbb{Z}\Rightarrow{x}\in\mathbb{Z} \\ $$$$\mathrm{let}\:{k},\:{m},\:{n}\:\in\mathbb{Z} \\ $$$$\left[\frac{{x}}{\mathrm{2}}\right]=\begin{cases}{\frac{{x}}{\mathrm{2}};\:{x}=\mathrm{2}{m}}\\{\frac{{x}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}};\:{x}=\mathrm{2}{m}+\mathrm{1}}\end{cases} \\ $$$$\left[\frac{\mathrm{2}{x}}{\mathrm{3}}\right]=\begin{cases}{\frac{\mathrm{2}{x}}{\mathrm{3}};\:{x}=\mathrm{3}{n}}\\{\frac{\mathrm{2}{x}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}};\:{x}=\mathrm{3}{n}+\mathrm{2}}\\{\frac{\mathrm{2}{x}}{\mathrm{3}}−\frac{\mathrm{2}}{\mathrm{3}};\:\mathrm{2}{x}=\mathrm{3}{n}+\mathrm{1}}\end{cases} \\ $$$$\left(\mathrm{1}\right)\:{x}=\mathrm{2}{m}=\mathrm{3}{n}\:\Rightarrow\:\:{x}=\mathrm{6}{k} \\ $$$$\:\:\:\:\:\left[\frac{\mathrm{6}{k}}{\mathrm{2}}\left[+\left[\frac{\mathrm{12}{k}}{\mathrm{3}}\right]=\mathrm{7}{k}\right.\right. \\ $$$$\:\:\:\:\:\mathrm{7}{k}=\mathrm{6}{k}\:\Rightarrow\:{k}=\mathrm{0}\:\Rightarrow\:{x}=\mathrm{0}\:\bullet \\ $$$$\left(\mathrm{2}\right)\:{x}=\mathrm{2}{m}=\mathrm{3}{n}+\mathrm{2}\:\Rightarrow\:{x}=\mathrm{6}{k}+\mathrm{2} \\ $$$$\:\:\:\:\:\left[\frac{\mathrm{6}{k}+\mathrm{2}}{\mathrm{2}}\right]+\left[\frac{\mathrm{12}{k}+\mathrm{4}}{\mathrm{3}}\right]=\mathrm{7}{k}+\mathrm{2} \\ $$$$\:\:\:\:\:\mathrm{7}{k}+\mathrm{2}=\mathrm{6}{k}+\mathrm{2}\:\Rightarrow\:{k}=\mathrm{0}\:\Rightarrow\:{x}=\mathrm{2}\:\bullet \\ $$$$\left(\mathrm{3}\right)\:{x}=\mathrm{2}{m}=\mathrm{3}{n}+\mathrm{1}\:\Rightarrow\:{x}=\mathrm{6}{k}+\mathrm{4} \\ $$$$\:\:\:\:\:\left[\frac{\mathrm{6}{k}+\mathrm{4}}{\mathrm{2}}\right]+\left[\frac{\mathrm{12}{k}+\mathrm{8}}{\mathrm{3}}\right]=\mathrm{7}{k}+\mathrm{4} \\ $$$$\:\:\:\:\:\mathrm{7}{k}+\mathrm{4}=\mathrm{6}{k}+\mathrm{4}\:\Rightarrow\:{k}=\mathrm{0}\:\Rightarrow\:{x}=\mathrm{4}\:\bullet \\ $$$$\left(\mathrm{4}\right)\:{x}=\mathrm{2}{m}+\mathrm{1}=\mathrm{3}{n}\:\Rightarrow\:{x}=\mathrm{6}{k}+\mathrm{3} \\ $$$$\:\:\:\:\:\left[\frac{\mathrm{6}{k}+\mathrm{3}}{\mathrm{2}}\right]+\left[\frac{\mathrm{12}{k}+\mathrm{6}}{\mathrm{3}}\right]=\mathrm{7}{k}+\mathrm{3} \\ $$$$\:\:\:\:\:\mathrm{7}{k}+\mathrm{3}=\mathrm{6}{k}+\mathrm{3}\:\Rightarrow\:{k}=\mathrm{0}\:\Rightarrow\:{x}=\mathrm{3}\:\bullet \\ $$$$\left(\mathrm{5}\right)\:{x}=\mathrm{2}{m}+\mathrm{1}=\mathrm{3}{n}+\mathrm{2}\:\Rightarrow\:{x}=\mathrm{6}{k}+\mathrm{5} \\ $$$$\:\:\:\:\:\left[\frac{\mathrm{6}{k}+\mathrm{5}}{\mathrm{2}}\right]+\left[\frac{\mathrm{12}{k}+\mathrm{10}}{\mathrm{3}}\right]=\mathrm{7}{k}+\mathrm{5} \\ $$$$\:\:\:\:\:\mathrm{7}{k}+\mathrm{5}=\mathrm{6}{k}+\mathrm{5}\:\Rightarrow\:{k}=\mathrm{0}\:\Rightarrow\:{x}=\mathrm{5}\:\bullet \\ $$$$\left(\mathrm{6}\right)\:{x}=\mathrm{2}{m}+\mathrm{1}=\mathrm{3}{n}+\mathrm{1}\:\Rightarrow\:{x}=\mathrm{6}{k}+\mathrm{1} \\ $$$$\:\:\:\:\:\left[\frac{\mathrm{6}{k}+\mathrm{1}}{\mathrm{2}}\right]+\left[\frac{\mathrm{12}{k}+\mathrm{2}}{\mathrm{3}}\right]=\mathrm{7}{k} \\ $$$$\:\:\:\:\:\mathrm{7}{k}=\mathrm{6}{k}+\mathrm{1}\:\Rightarrow\:{k}=\mathrm{1}\:\Rightarrow\:{x}=\mathrm{7}\:\bullet \\ $$$$ \\ $$$$\mathrm{answer}\:\mathrm{is}\:{x}\in\left\{\mathrm{0},\:\mathrm{2},\:\mathrm{3},\:\mathrm{4},\:\mathrm{5},\:\mathrm{7}\right\} \\ $$
Commented by M±th+et£s last updated on 26/Mar/20
thank you sir god bless you
$${thank}\:{you}\:{sir}\:{god}\:{bless}\:{you} \\ $$

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