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Question Number 119440 by mathocean1 last updated on 24/Oct/20

Determine the integer which can be written: N=xyz^(−) ^7 =zyx^(−) ^(11)

$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{integer}\:\mathrm{which}\:\mathrm{can}\: \\ $$$$\mathrm{be}\:\mathrm{written}:\:\mathrm{N}=\overline {\mathrm{xyz}}\:^{\mathrm{7}} =\overline {\mathrm{zyx}}\:^{\mathrm{11}} \\ $$

Commented by mathocean1 last updated on 24/Oct/20

Please can you detail sir

$$\mathrm{Please}\:\mathrm{can}\:\mathrm{you}\:\mathrm{detail}\:\mathrm{sir} \\ $$

Answered by mr W last updated on 25/Oct/20

x,y,z are digits of an integer of base 7, ⇒0≤x,y,z≤6 besides, x≥1, z≥1. (xyz)_7 =49x+7y+z (zyx)_(11) =121z+11y+x ⇒49x+7y+z=121z+11y+x ⇒48x−4y=120z ⇒12x−y=30z y must be a multiple of 6, say y=6Y ⇒2x−Y=5z Y=0 or 1 with Y=0: 2x=5z ⇒x=5,z=2 with Y=1: 2x−1=5z ⇒x=3, z=1 the solutions are: x=5, Y=0 (i.e. y=0), z=2 ⇒N=(502)_7 =(205)_(11) =247 x=3, Y=1 (i.e. y=6), z=1 ⇒N=(361)_7 =(163)_(11) =190

$${x},{y},{z}\:{are}\:{digits}\:{of}\:{an}\:{integer}\:{of}\:{base}\:\mathrm{7}, \\ $$$$\Rightarrow\mathrm{0}\leqslant{x},{y},{z}\leqslant\mathrm{6} \\ $$$${besides},\:{x}\geqslant\mathrm{1},\:{z}\geqslant\mathrm{1}. \\ $$$$ \\ $$$$\left({xyz}\right)_{\mathrm{7}} =\mathrm{49}{x}+\mathrm{7}{y}+{z} \\ $$$$\left({zyx}\right)_{\mathrm{11}} =\mathrm{121}{z}+\mathrm{11}{y}+{x} \\ $$$$\Rightarrow\mathrm{49}{x}+\mathrm{7}{y}+{z}=\mathrm{121}{z}+\mathrm{11}{y}+{x} \\ $$$$\Rightarrow\mathrm{48}{x}−\mathrm{4}{y}=\mathrm{120}{z} \\ $$$$\Rightarrow\mathrm{12}{x}−{y}=\mathrm{30}{z} \\ $$$${y}\:{must}\:{be}\:{a}\:{multiple}\:{of}\:\mathrm{6},\:{say}\:{y}=\mathrm{6}{Y} \\ $$$$\Rightarrow\mathrm{2}{x}−{Y}=\mathrm{5}{z} \\ $$$${Y}=\mathrm{0}\:{or}\:\mathrm{1} \\ $$$${with}\:{Y}=\mathrm{0}: \\ $$$$\mathrm{2}{x}=\mathrm{5}{z} \\ $$$$\Rightarrow{x}=\mathrm{5},{z}=\mathrm{2} \\ $$$${with}\:{Y}=\mathrm{1}: \\ $$$$\mathrm{2}{x}−\mathrm{1}=\mathrm{5}{z} \\ $$$$\Rightarrow{x}=\mathrm{3},\:{z}=\mathrm{1} \\ $$$$ \\ $$$${the}\:{solutions}\:{are}: \\ $$$${x}=\mathrm{5},\:{Y}=\mathrm{0}\:\left({i}.{e}.\:{y}=\mathrm{0}\right),\:{z}=\mathrm{2} \\ $$$$\Rightarrow{N}=\left(\mathrm{502}\right)_{\mathrm{7}} =\left(\mathrm{205}\right)_{\mathrm{11}} =\mathrm{247} \\ $$$${x}=\mathrm{3},\:{Y}=\mathrm{1}\:\left({i}.{e}.\:{y}=\mathrm{6}\right),\:{z}=\mathrm{1} \\ $$$$\Rightarrow{N}=\left(\mathrm{361}\right)_{\mathrm{7}} =\left(\mathrm{163}\right)_{\mathrm{11}} =\mathrm{190} \\ $$

Commented by mathocean1 last updated on 25/Oct/20

Thank you sir.

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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