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Question Number 168956 by mnjuly1970 last updated on 22/Apr/22
solve ⌊ x ⌋ + ⌊ (x/6) ⌋ =(2/3) x
solve⌊x⌋+⌊x6⌋=23x
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
letx=6n+k+rwheren∈Zk=0,1,2,3,4,5r∈[0,1)⇒⌊x⌋+⌊x6⌋=6n+k+n=23(6n+k+r)⇒7n+k=4n+23k+23r⇒3n=2r−k3⇒9n=2r−kr∈[0,1)⇒2r∈[0,2)⇒2r−k∈[−5,2)⇒9n∈[−5,2)butalson∈Z⇒9n∈Z⇒9n∈Z∩[−5,2)={−5,−4,…,0,1}theonlypossiblewholenis0⇒n=0⇒k=2r∈[0,2)andk∈{0,1,2,3,4,5}⇒(k,r)∈{(0,0),(1,0.5)}⇒x=0andx=1.5
Answered by MJS_new last updated on 22/Apr/22
x=0∨x=(3/2) sorry no path, just by intuition
x=0∨x=32sorrynopath,justbyintuition
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
[x]+[x6]=2x3∈Z⇒x∈32Z={3k2,k∈Z}⇒[1.5k]+[0.25k]=k⇒getk=4i+ji∈Zj=0,1,2,3⇒6i+[1.5j]+i=4i+j⇒i=j−[1.5j]3∈Z⇒forj=0⇒i=0,j=1⇒i=0j=2or3⇒i=−13∉Z⇒x=32k=6i+1.5j=0or1.5
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=32k(k∈Z)k+⌊k2⌋+⌊k4⌋=k⌊k2⌋=−⌊k4⌋,4r⩽k<4r+4(r∈Z)k=(1)4r,4r+(2)1,4r+(3)2,4r+(4)3∴(1)::2r=−r⇒r=0⇒x=0✓(2)::2r=−r⇒r=0⇒x=32✓(3)::2r+1=−r⇒r=−13⇒k=23∉Z(4)::2r+1=−r⇒r=−13⇒k=53∉Zx∈{0,32}

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