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Question Number 102588 by bobhans last updated on 10/Jul/20
solve 109x +103y = 5 for x,y are integer
$${solve}\:\mathrm{109}{x}\:+\mathrm{103}{y}\:=\:\mathrm{5}\:{for}\:{x},{y}\:{are}\:{integer} \\ $$
Commented by mr W last updated on 10/Jul/20
see Q44819
$${see}\:{Q}\mathrm{44819} \\ $$
Commented by mr W last updated on 10/Jul/20
see also Q19198
$${see}\:{also}\:{Q}\mathrm{19198} \\ $$
Commented by bobhans last updated on 10/Jul/20
coll sir
$${coll}\:{sir} \\ $$
Answered by 1549442205 last updated on 10/Jul/20
y=((5−109x)/(103))=((5−6x)/(103))−x.Put ((5−6x)/(103))=a(a∈Z) ⇒5−6x=103a⇔x=((5−103a)/6)=−17a+((5−a)/6) Put ((5−a)/6)=b⇒a=5−6b.From this we get x=−17a+b=103b−85,y=−x+a =−109b+90.Thus,the roots of eqs be { ((x=103b−85)),((y=−109b+90)) :}(b∈Z)
$$\mathrm{y}=\frac{\mathrm{5}−\mathrm{109x}}{\mathrm{103}}=\frac{\mathrm{5}−\mathrm{6x}}{\mathrm{103}}−\mathrm{x}.\mathrm{Put}\:\frac{\mathrm{5}−\mathrm{6x}}{\mathrm{103}}=\mathrm{a}\left(\mathrm{a}\in\mathbb{Z}\right) \\ $$$$\Rightarrow\mathrm{5}−\mathrm{6x}=\mathrm{103a}\Leftrightarrow\mathrm{x}=\frac{\mathrm{5}−\mathrm{103a}}{\mathrm{6}}=−\mathrm{17a}+\frac{\mathrm{5}−\mathrm{a}}{\mathrm{6}} \\ $$$$\mathrm{Put}\:\frac{\mathrm{5}−\mathrm{a}}{\mathrm{6}}=\mathrm{b}\Rightarrow\mathrm{a}=\mathrm{5}−\mathrm{6b}.\mathrm{From}\:\mathrm{this}\:\mathrm{we} \\ $$$$\mathrm{get}\:\mathrm{x}=−\mathrm{17a}+\mathrm{b}=\mathrm{103b}−\mathrm{85},\mathrm{y}=−\mathrm{x}+\mathrm{a} \\ $$$$=−\mathrm{109b}+\mathrm{90}.\mathrm{Thus},\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{eqs}\:\mathrm{be} \\ $$$$\begin{cases}{\boldsymbol{\mathrm{x}}=\mathrm{103}\boldsymbol{\mathrm{b}}−\mathrm{85}}\\{\boldsymbol{\mathrm{y}}=−\mathrm{109}\boldsymbol{\mathrm{b}}+\mathrm{90}}\end{cases}\left(\boldsymbol{\mathrm{b}}\in\mathbb{Z}\right) \\ $$$$ \\ $$
Answered by bemath last updated on 10/Jul/20
{ ((109=1(103)+6)),((103=17(6)+1)) :} { ((103−17(6)=1)),((103−17(109−103) =1)) :} ⇔109(−17)+103(18)=1...(×5) 109(−85)+103(90) = 5 we get generall solution x=−85 +103n y = 90−109n , n∈Z
$$\begin{cases}{\mathrm{109}=\mathrm{1}\left(\mathrm{103}\right)+\mathrm{6}}\\{\mathrm{103}=\mathrm{17}\left(\mathrm{6}\right)+\mathrm{1}}\end{cases} \\ $$$$\begin{cases}{\mathrm{103}−\mathrm{17}\left(\mathrm{6}\right)=\mathrm{1}}\\{\mathrm{103}−\mathrm{17}\left(\mathrm{109}−\mathrm{103}\right)\:=\mathrm{1}}\end{cases} \\ $$$$\Leftrightarrow\mathrm{109}\left(−\mathrm{17}\right)+\mathrm{103}\left(\mathrm{18}\right)=\mathrm{1}…\left(×\mathrm{5}\right) \\ $$$$\mathrm{109}\left(−\mathrm{85}\right)+\mathrm{103}\left(\mathrm{90}\right)\:=\:\mathrm{5} \\ $$$${we}\:{get}\:{generall}\:{solution} \\ $$$${x}=−\mathrm{85}\:+\mathrm{103}{n} \\ $$$${y}\:=\:\mathrm{90}−\mathrm{109}{n}\:,\:{n}\in\mathbb{Z} \\ $$
Answered by PRITHWISH SEN 2 last updated on 10/Jul/20
109x+103y=5 It is a diophantine equation ∵ gcd(109,103)=1 then there exists two integers u and v such that 109u+103v=1 here u=−85 amd v=90 then 109x+103y=−109.85+103.90 ⇒109(x+85)=−103(y−90) ((x+85)/(−103)) = ((y−90)/(109)) = t (integer) ∵ gcd(109,103)=1 then103∣(x+85) and 109∣(y−90) ∴ x =−103t−85 y = 109t+90
$$\mathrm{109x}+\mathrm{103y}=\mathrm{5}\:\:\:\:\boldsymbol{\mathrm{It}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{diophantine}}\:\boldsymbol{\mathrm{equation}} \\ $$$$\because\:\mathrm{gcd}\left(\mathrm{109},\mathrm{103}\right)=\mathrm{1} \\ $$$$\mathrm{then}\:\mathrm{there}\:\mathrm{exists}\:\mathrm{two}\:\mathrm{integers}\:\mathrm{u}\:\mathrm{and}\:\mathrm{v}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{109u}+\mathrm{103v}=\mathrm{1} \\ $$$$\mathrm{here}\:\mathrm{u}=−\mathrm{85}\:\mathrm{amd}\:\mathrm{v}=\mathrm{90} \\ $$$$\mathrm{then}\:\mathrm{109x}+\mathrm{103y}=−\mathrm{109}.\mathrm{85}+\mathrm{103}.\mathrm{90} \\ $$$$\Rightarrow\mathrm{109}\left(\mathrm{x}+\mathrm{85}\right)=−\mathrm{103}\left(\mathrm{y}−\mathrm{90}\right) \\ $$$$\frac{\mathrm{x}+\mathrm{85}}{−\mathrm{103}}\:=\:\frac{\mathrm{y}−\mathrm{90}}{\mathrm{109}}\:\:=\:\mathrm{t}\:\left(\boldsymbol{\mathrm{integer}}\right)\: \\ $$$$\:\because\:\mathrm{gcd}\left(\mathrm{109},\mathrm{103}\right)=\mathrm{1}\:\mathrm{then103}\mid\left(\mathrm{x}+\mathrm{85}\right)\:\mathrm{and}\:\mathrm{109}\mid\left(\mathrm{y}−\mathrm{90}\right) \\ $$$$\therefore\:\boldsymbol{\mathrm{x}}\:=−\mathrm{103}\boldsymbol{\mathrm{t}}−\mathrm{85} \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{y}}\:=\:\mathrm{109}\boldsymbol{\mathrm{t}}+\mathrm{90} \\ $$$$ \\ $$
Answered by bobhans last updated on 10/Jul/20
well has 2 solution??
$${well}\:{has}\:\mathrm{2}\:{solution}??\: \\ $$

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