solve-109x-103y-5-for-x-y-are-integer-

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

s o l v e 109 x + 103 y = 5 f o r x, y a r e i n t e g e r

Commented by mr W last updated on 10/Jul/20

s e e Q 44819

Commented by mr W last updated on 10/Jul/20

s e e a l s o Q 19198

Commented by bobhans last updated on 10/Jul/20

c o l l s i r

Answered by 1549442205 last updated on 10/Jul/20

y = \frac{5 - 109 x}{103} = \frac{5 - 6 x}{103} - x . Put \frac{5 - 6 x}{103} = a (a \in Z)

\Rightarrow 5 - 6 x = 103 a \Leftrightarrow x = \frac{5 - 103 a}{6} = - 17 a + \frac{5 - a}{6}

Put \frac{5 - a}{6} = b \Rightarrow a = 5 - 6 b . From this we

get x = - 17 a + b = 103 b - 85, y = - x + a

= - 109 b + 90 . Thus, the roots of eqs be

{\begin{cases} x = 103 b - 85 \\ y = - 109 b + 90 \end{cases} (b \in Z)

Answered by bemath last updated on 10/Jul/20

{\begin{cases} 109 = 1 (103) + 6 \\ 103 = 17 (6) + 1 \end{cases}

{\begin{cases} 103 - 17 (6) = 1 \\ 103 - 17 (109 - 103) = 1 \end{cases}

\Leftrightarrow 109 (- 17) + 103 (18) = 1 \dots (\times 5)

109 (- 85) + 103 (90) = 5

w e g e t g e n e r a l l s o l u t i o n

x = - 85 + 103 n

y = 90 - 109 n, n \in Z

Answered by PRITHWISH SEN 2 last updated on 10/Jul/20

109 x + 103 y = 5 It is a diophantine equation

∵ \gcd (109, 103) = 1

then there exists two integers u and v such that

109 u + 103 v = 1

here u = - 85 amd v = 90

then 109 x + 103 y = - 109.85 + 103.90

\Rightarrow 109 (x + 85) = - 103 (y - 90)

\frac{x + 85}{- 103} = \frac{y - 90}{109} = t (integer)

∵ \gcd (109, 103) = 1 then 103 ∣ (x + 85) and 109 ∣ (y - 90)

∴ x = - 103 t - 85

y = 109 t + 90

Answered by bobhans last updated on 10/Jul/20

w e l l h a s 2 s o l u t i o n ? ?

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