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Question Number 192172 by mehdee42 last updated on 10/May/23

Q1 ∴ x=<1a_1 a_2 ...a_n >∈N & y=<a_1 a_2 ...a_n 1>∈N if y=3x then , find the smallest value of x Q2 ∴ with the above conditions ,what other values can be placed besides the number ♮ 1 ε

$${Q}\mathrm{1}\:\therefore\:\:{x}=<\mathrm{1}{a}_{\mathrm{1}} {a}_{\mathrm{2}} ...{a}_{{n}} >\in\mathbb{N}\:\:\&\:\:{y}=<{a}_{\mathrm{1}} {a}_{\mathrm{2}} ...{a}_{{n}} \mathrm{1}>\in\mathbb{N} \\ $$ $${if}\:\:{y}=\mathrm{3}{x}\:\:{then}\:\:,\:{find}\:{the}\:{smallest}\: \\ $$ $${value}\:{of}\:\:{x} \\ $$ $${Q}\mathrm{2}\:\therefore\:{with}\:{the}\:{above}\:{conditions}\:,{what}\:{other}\:{values}\: \\ $$ $${can}\:{be}\:{placed}\:\:{besides}\:{the}\:{number}\:\natural\:\mathrm{1}\:\varepsilon\: \\ $$

Answered by AST last updated on 10/May/23

1) x=10^n +((y−1)/(10)); y=3x⇒x=10^n +((3x−1)/(10)) ⇒10x=10^(n+1) +3x−1⇒7x=10^(n+1) −1 10^(n+1) −1≡0(mod 7)⇒3^(n+1) ≡1(mod 7)⇒3^n ≡^(7) 5 ord_3 (7)=6;3^5 ≡^7 5; φ(7)=6⇒n=6k+5; min(n)=5 ⇒min(x)=((10^6 −1)/7)=142857 2)x=p10^n +((3x−p)/(10))⇒10x=p10^(n+1) +3x−p ⇒7x=p(10^(n+1) −1) The first digit of ((p(10^(n+1) −1))/7) should be p This is only true for p=1,2 ⇒p=1 and 2 always work when n=6k+5. Hence,only 2 can be placed besides the number ♮1ε.

$$\left.\mathrm{1}\right)\:{x}=\mathrm{10}^{{n}} +\frac{{y}−\mathrm{1}}{\mathrm{10}};\:\:{y}=\mathrm{3}{x}\Rightarrow{x}=\mathrm{10}^{{n}} +\frac{\mathrm{3}{x}−\mathrm{1}}{\mathrm{10}} \\ $$ $$\Rightarrow\mathrm{10}{x}=\mathrm{10}^{{n}+\mathrm{1}} +\mathrm{3}{x}−\mathrm{1}\Rightarrow\mathrm{7}{x}=\mathrm{10}^{{n}+\mathrm{1}} −\mathrm{1} \\ $$ $$\mathrm{10}^{{n}+\mathrm{1}} −\mathrm{1}\equiv\mathrm{0}\left({mod}\:\mathrm{7}\right)\Rightarrow\mathrm{3}^{{n}+\mathrm{1}} \equiv\mathrm{1}\left({mod}\:\mathrm{7}\right)\Rightarrow\mathrm{3}^{{n}} \overset{\mathrm{7}} {\equiv}\mathrm{5} \\ $$ $${ord}_{\mathrm{3}} \left(\mathrm{7}\right)=\mathrm{6};\mathrm{3}^{\mathrm{5}} \overset{\mathrm{7}} {\equiv}\mathrm{5};\:\phi\left(\mathrm{7}\right)=\mathrm{6}\Rightarrow{n}=\mathrm{6}{k}+\mathrm{5};\:{min}\left({n}\right)=\mathrm{5} \\ $$ $$\Rightarrow{min}\left({x}\right)=\frac{\mathrm{10}^{\mathrm{6}} −\mathrm{1}}{\mathrm{7}}=\mathrm{142857} \\ $$ $$ \\ $$ $$\left.\mathrm{2}\right){x}={p}\mathrm{10}^{{n}} +\frac{\mathrm{3}{x}−{p}}{\mathrm{10}}\Rightarrow\mathrm{10}{x}={p}\mathrm{10}^{{n}+\mathrm{1}} +\mathrm{3}{x}−{p} \\ $$ $$\Rightarrow\mathrm{7}{x}={p}\left(\mathrm{10}^{{n}+\mathrm{1}} −\mathrm{1}\right) \\ $$ $${The}\:{first}\:{digit}\:{of}\:\:\frac{{p}\left(\mathrm{10}^{{n}+\mathrm{1}} −\mathrm{1}\right)}{\mathrm{7}}\:{should}\:{be}\:{p} \\ $$ $${This}\:{is}\:{only}\:{true}\:{for}\:{p}=\mathrm{1},\mathrm{2} \\ $$ $$\Rightarrow{p}=\mathrm{1}\:{and}\:\mathrm{2}\:{always}\:{work}\:{when}\:{n}=\mathrm{6}{k}+\mathrm{5}. \\ $$ $${Hence},{only}\:\mathrm{2}\:{can}\:{be}\:{placed}\:{besides}\:{the}\:{number} \\ $$ $$\natural\mathrm{1}\varepsilon. \\ $$

Commented bymehdee42 last updated on 10/May/23

very nice Sir for p=2 ; the lowest value x=285714

$${very}\:{nice}\:{Sir} \\ $$ $${for}\:{p}=\mathrm{2}\:;\:{the}\:{lowest}\:{value}\: \\ $$ $${x}=\mathrm{285714} \\ $$

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