solve ⌊ x ⌋ + ⌊ (x/6) ⌋ =(2/3) x


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Question Number 168956 by mnjuly1970 last updated on 22/Apr/22

$s o l v e$ $⌊ x ⌋ + ⌊ \frac{x}{6} ⌋ = \frac{2}{3} x$
	Answered by aleks041103 last updated on 22/Apr/22

	$l e t x = 6 n + k + r$ $w h e r e$ $n \in Z$ $k = 0, 1, 2, 3, 4, 5$ $r \in [0, 1)$ $\Rightarrow ⌊ x ⌋ + ⌊ \frac{x}{6} ⌋ = 6 n + k + n = \frac{2}{3} (6 n + k + r)$ $\Rightarrow 7 n + k = 4 n + \frac{2}{3} k + \frac{2}{3} r$ $\Rightarrow 3 n = \frac{2 r - k}{3}$ $\Rightarrow 9 n = 2 r - k$ $r \in [0, 1) \Rightarrow 2 r \in [0, 2)$ $\Rightarrow 2 r - k \in [- 5, 2)$ $\Rightarrow 9 n \in [- 5, 2)$ $b u t a l s o n \in Z \Rightarrow 9 n \in Z$ $\Rightarrow 9 n \in Z \cap [- 5, 2) = {- 5, - 4, . . ., 0, 1}$ $t h e o n l y p o s s i b l e w h o l e n i s 0$ $\Rightarrow n = 0$ $\Rightarrow k = 2 r \in [0, 2)$ $a n d k \in {0, 1, 2, 3, 4, 5}$ $\Rightarrow (k, r) \in {(0, 0), (1, 0.5)}$ $\Rightarrow x = 0 a n d x = 1.5$
	Answered by MJS_new last updated on 22/Apr/22

	$x = 0 \lor x = \frac{3}{2}$ $sorry no path, just by intuition$
	Answered by mahdipoor last updated on 22/Apr/22

	$[x] + [\frac{x}{6}] = \frac{2 x}{3} \in Z \Rightarrow x \in \frac{3}{2} Z = {\frac{3 k}{2}, k \in Z}$ $\Rightarrow [1.5 k] + [0.25 k] = k$ $\Rightarrow g e t k = 4 i + j i \in Z j = 0, 1, 2, 3$ $\Rightarrow 6 i + [1.5 j] + i = 4 i + j \Rightarrow i = \frac{j - [1.5 j]}{3} \in Z$ $\Rightarrow f o r j = 0 \Rightarrow i = 0, j = 1 \Rightarrow i = 0$ $j = 2 o r 3 \Rightarrow i = \frac{- 1}{3} \notin Z$ $\Rightarrow x = \frac{3}{2} k = 6 i + 1.5 j = 0 o r 1.5$
	Answered by mnjuly1970 last updated on 22/Apr/22
	$x= (3/2) k (k ∈ Z ) k + ⌊ (k/2) ⌋+ ⌊ (k/4)⌋ = k ⌊ (k/2) ⌋ = −⌊(k/4) ⌋ , 4 r≤k < 4r +4 ( r ∈ Z ) k =^((1)) 4r , 4r+^((2)) 1 , 4r +^((3)) 2 , 4r +^((4)) 3 ∴ (1):: 2 r = −r ⇒ r =0 ⇒ x=0✓ ( 2) :: 2 r = −r ⇒ r = 0 ⇒ x= (3/2) ✓ (3) :: 2r +1 = −r ⇒ r=((−1)/3) ⇒ k=(2/3) ∉ Z (4 ):: 2r +1 = −r ⇒ r =((−1)/3) ⇒ k =(5/3) ∉ Z x∈ { 0 , (3/2) }$
	$x = \frac{3}{2} k (k \in Z)$ $k + ⌊ \frac{k}{2} ⌋ + ⌊ \frac{k}{4} ⌋ = k$ $⌊ \frac{k}{2} ⌋ = - ⌊ \frac{k}{4} ⌋, 4 r ⩽ k < 4 r + 4 (r \in Z)$ $k \overset{(1)}{=} 4 r, 4 r \overset{(2)}{+} 1, 4 r \overset{(3)}{+} 2, 4 r \overset{(4)}{+} 3$ $∴ (1) :: 2 r = - r \Rightarrow r = 0 \Rightarrow x = 0 ✓$ $(2) :: 2 r = - r \Rightarrow r = 0 \Rightarrow x = \frac{3}{2} ✓$ $(3) :: 2 r + 1 = - r \Rightarrow r = \frac{- 1}{3} \Rightarrow k = \frac{2}{3} \notin Z$ $(4) :: 2 r + 1 = - r \Rightarrow r = \frac{- 1}{3} \Rightarrow k = \frac{5}{3} \notin Z$ $x \in {0, \frac{3}{2}}$
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