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Question Number 168956 by mnjuly1970 last updated on 22/Apr/22
solve ⌊ x ⌋ + ⌊ (x/6) ⌋ =(2/3) x
$$ \\ $$$$\:\:\:\:{solve} \\ $$$$\:\:\lfloor\:{x}\:\rfloor\:+\:\lfloor\:\frac{{x}}{\mathrm{6}}\:\rfloor\:=\frac{\mathrm{2}}{\mathrm{3}}\:{x} \\ $$$$ \\ $$
Answered by aleks041103 last updated on 22/Apr/22
let x=6n+k+r where n∈Z k=0,1,2,3,4,5 r∈[0,1) ⇒⌊x⌋+⌊(x/6)⌋=6n+k+n=(2/3)(6n+k+r) ⇒7n+k=4n+(2/3)k+(2/3)r ⇒3n=((2r−k)/3) ⇒9n=2r−k r∈[0,1)⇒2r∈[0,2) ⇒2r−k∈[−5,2) ⇒9n∈[−5,2) but also n∈Z⇒9n∈Z ⇒9n∈Z∩[−5,2)={−5,−4,...,0,1} the only possible whole n is 0 ⇒n=0 ⇒k=2r∈[0,2) and k∈{0,1,2,3,4,5} ⇒(k,r)∈{(0,0),(1,0.5)} ⇒x=0 and x=1.5
$${let}\:{x}=\mathrm{6}{n}+{k}+{r} \\ $$$${where} \\ $$$${n}\in\mathbb{Z} \\ $$$${k}=\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5} \\ $$$${r}\in\left[\mathrm{0},\mathrm{1}\right) \\ $$$$\Rightarrow\lfloor{x}\rfloor+\lfloor\frac{{x}}{\mathrm{6}}\rfloor=\mathrm{6}{n}+{k}+{n}=\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{6}{n}+{k}+{r}\right) \\ $$$$\Rightarrow\mathrm{7}{n}+{k}=\mathrm{4}{n}+\frac{\mathrm{2}}{\mathrm{3}}{k}+\frac{\mathrm{2}}{\mathrm{3}}{r} \\ $$$$\Rightarrow\mathrm{3}{n}=\frac{\mathrm{2}{r}−{k}}{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{9}{n}=\mathrm{2}{r}−{k} \\ $$$${r}\in\left[\mathrm{0},\mathrm{1}\right)\Rightarrow\mathrm{2}{r}\in\left[\mathrm{0},\mathrm{2}\right) \\ $$$$\Rightarrow\mathrm{2}{r}−{k}\in\left[−\mathrm{5},\mathrm{2}\right) \\ $$$$\Rightarrow\mathrm{9}{n}\in\left[−\mathrm{5},\mathrm{2}\right) \\ $$$${but}\:{also}\:{n}\in\mathbb{Z}\Rightarrow\mathrm{9}{n}\in\mathbb{Z} \\ $$$$\Rightarrow\mathrm{9}{n}\in\mathbb{Z}\cap\left[−\mathrm{5},\mathrm{2}\right)=\left\{−\mathrm{5},−\mathrm{4},…,\mathrm{0},\mathrm{1}\right\} \\ $$$${the}\:{only}\:{possible}\:{whole}\:{n}\:{is}\:\mathrm{0} \\ $$$$\Rightarrow{n}=\mathrm{0} \\ $$$$\Rightarrow{k}=\mathrm{2}{r}\in\left[\mathrm{0},\mathrm{2}\right) \\ $$$${and}\:{k}\in\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5}\right\} \\ $$$$\Rightarrow\left({k},{r}\right)\in\left\{\left(\mathrm{0},\mathrm{0}\right),\left(\mathrm{1},\mathrm{0}.\mathrm{5}\right)\right\} \\ $$$$\Rightarrow{x}=\mathrm{0}\:{and}\:{x}=\mathrm{1}.\mathrm{5} \\ $$
Answered by MJS_new last updated on 22/Apr/22
x=0∨x=(3/2) sorry no path, just by intuition
$${x}=\mathrm{0}\vee{x}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{sorry}\:\mathrm{no}\:\mathrm{path},\:\mathrm{just}\:\mathrm{by}\:\mathrm{intuition} \\ $$
Answered by mahdipoor last updated on 22/Apr/22
[x]+[(x/6)]=((2x)/3)∈Z ⇒ x∈(3/2)Z={((3k)/2),k∈Z} ⇒[1.5k]+[0.25k]=k ⇒get k=4i+j i∈Z j=0,1,2,3 ⇒6i+[1.5j]+i=4i+j⇒i=((j−[1.5j])/3)∈Z ⇒for j=0 ⇒ i=0 , j=1 ⇒ i=0 j=2 or 3 ⇒ i=((−1)/3)∉Z ⇒x=(3/2)k=6i+1.5j=0 or 1.5
$$\left[{x}\right]+\left[\frac{{x}}{\mathrm{6}}\right]=\frac{\mathrm{2}{x}}{\mathrm{3}}\in\mathrm{Z}\:\Rightarrow\:{x}\in\frac{\mathrm{3}}{\mathrm{2}}\mathrm{Z}=\left\{\frac{\mathrm{3}{k}}{\mathrm{2}},{k}\in\mathrm{Z}\right\} \\ $$$$\Rightarrow\left[\mathrm{1}.\mathrm{5}{k}\right]+\left[\mathrm{0}.\mathrm{25}{k}\right]={k}\: \\ $$$$\Rightarrow{get}\:{k}=\mathrm{4}{i}+{j}\:\:\:\:{i}\in\mathrm{Z}\:\:\:{j}=\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3} \\ $$$$\Rightarrow\mathrm{6}{i}+\left[\mathrm{1}.\mathrm{5}{j}\right]+{i}=\mathrm{4}{i}+{j}\Rightarrow{i}=\frac{{j}−\left[\mathrm{1}.\mathrm{5}{j}\right]}{\mathrm{3}}\in\mathrm{Z} \\ $$$$\Rightarrow{for}\:{j}=\mathrm{0}\:\Rightarrow\:{i}=\mathrm{0}\:\:\:,\:\:{j}=\mathrm{1}\:\Rightarrow\:{i}=\mathrm{0}\: \\ $$$$\:\:\:\:\:\:\:\:\:{j}=\mathrm{2}\:{or}\:\mathrm{3}\:\Rightarrow\:{i}=\frac{−\mathrm{1}}{\mathrm{3}}\notin\mathrm{Z} \\ $$$$\Rightarrow{x}=\frac{\mathrm{3}}{\mathrm{2}}{k}=\mathrm{6}{i}+\mathrm{1}.\mathrm{5}{j}=\mathrm{0}\:{or}\:\mathrm{1}.\mathrm{5} \\ $$$$ \\ $$
Answered by mnjuly1970 last updated on 22/Apr/22
x= (3/2) k (k ∈ Z ) k + ⌊ (k/2) ⌋+ ⌊ (k/4)⌋ = k ⌊ (k/2) ⌋ = −⌊(k/4) ⌋ , 4 r≤k < 4r +4 ( r ∈ Z ) k =^((1)) 4r , 4r+^((2)) 1 , 4r +^((3)) 2 , 4r +^((4)) 3 ∴ (1):: 2 r = −r ⇒ r =0 ⇒ x=0✓ ( 2) :: 2 r = −r ⇒ r = 0 ⇒ x= (3/2) ✓ (3) :: 2r +1 = −r ⇒ r=((−1)/3) ⇒ k=(2/3) ∉ Z (4 ):: 2r +1 = −r ⇒ r =((−1)/3) ⇒ k =(5/3) ∉ Z x∈ { 0 , (3/2) }
$$\:\:\:{x}=\:\frac{\mathrm{3}}{\mathrm{2}}\:{k}\:\:\:\left({k}\:\in\:\mathbb{Z}\:\right)\: \\ $$$$\:\:\:\:\:\:\:{k}\:+\:\lfloor\:\frac{{k}}{\mathrm{2}}\:\rfloor+\:\lfloor\:\frac{{k}}{\mathrm{4}}\rfloor\:=\:{k} \\ $$$$\:\:\:\:\:\:\:\:\:\:\lfloor\:\frac{{k}}{\mathrm{2}}\:\rfloor\:=\:−\lfloor\frac{{k}}{\mathrm{4}}\:\rfloor\:\:,\:\:\mathrm{4}\:{r}\leqslant{k}\:<\:\mathrm{4}{r}\:+\mathrm{4}\:\left(\:{r}\:\in\:\mathbb{Z}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:{k}\:\overset{\left(\mathrm{1}\right)} {=}\:\mathrm{4}{r}\:,\:\mathrm{4}{r}\overset{\left(\mathrm{2}\right)} {+}\mathrm{1}\:,\:\mathrm{4}{r}\:\overset{\left(\mathrm{3}\right)} {+}\mathrm{2}\:,\:\mathrm{4}{r}\:\overset{\left(\mathrm{4}\right)} {+}\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\therefore\:\:\:\left(\mathrm{1}\right)::\:\:\mathrm{2}\:{r}\:=\:−{r}\:\:\Rightarrow\:\:{r}\:=\mathrm{0}\:\:\Rightarrow\:{x}=\mathrm{0}\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\:\mathrm{2}\right)\:::\:\:\:\:\mathrm{2}\:{r}\:=\:−{r}\:\Rightarrow\:{r}\:=\:\mathrm{0}\:\Rightarrow\:{x}=\:\frac{\mathrm{3}}{\mathrm{2}}\:\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{3}\right)\:::\:\:\mathrm{2}{r}\:+\mathrm{1}\:=\:−{r}\:\Rightarrow\:{r}=\frac{−\mathrm{1}}{\mathrm{3}}\:\Rightarrow\:\:{k}=\frac{\mathrm{2}}{\mathrm{3}}\:\notin\:\mathbb{Z}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{4}\:\right)::\:\:\:\:\:\mathrm{2}{r}\:+\mathrm{1}\:=\:−{r}\:\Rightarrow\:{r}\:=\frac{−\mathrm{1}}{\mathrm{3}}\:\Rightarrow\:\:{k}\:=\frac{\mathrm{5}}{\mathrm{3}}\:\notin\:\mathbb{Z}\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\in\:\left\{\:\mathrm{0}\:,\:\frac{\mathrm{3}}{\mathrm{2}}\:\right\} \\ $$

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