Question Number 102588 by bobhans last updated on 10/Jul/20

solve 109x +103y = 5 for x,y are integer

Commented bymr W last updated on 10/Jul/20

see Q44819

Commented bymr W last updated on 10/Jul/20

see also Q19198

Commented bybobhans last updated on 10/Jul/20

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)

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

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

Answered by bobhans last updated on 10/Jul/20

well has 2 solution??