4x-2-mod-9-7x-2-mod-13-

Question Number 175320 by cortano1 last updated on 27/Aug/22

{\begin{cases} 4 x = 2 (m o d 9) \\ 7 x = 2 (m o d 13) \end{cases}

Commented by cortano1 last updated on 27/Aug/22

i g o t x = 23 + 117 k s i r

Commented by Rasheed.Sindhi last updated on 27/Aug/22

s i r x = 23 + 117 k {d o e s n}^{'} t s a t i s f y

7 x \equiv 2 (m o d 13)

Commented by cortano1 last updated on 27/Aug/22

o y e s . i a m t y p o 6 \times 13 = 68

Answered by Rasheed.Sindhi last updated on 27/Aug/22

4 x \equiv 2 + 2 \times 9 (m o d 9)

x \equiv 5 (m o d 9)

x = 5 + 9 m

7 x \equiv 2 (m o d 13) \Rightarrow 7 (5 + 9 m) \equiv 2 (m o d 13

35 + 63 m \equiv 2 (m o d 13)

63 m \equiv - 33 (m o d 13)

m = 10

x = 5 + 9 m = 5 + 9 (10) = 95

x = 95 + k \times LCM (9, 13); k \in Z

x = 95 + 117 k; k \in Z

Commented by Tawa11 last updated on 27/Aug/22

Great sir

Answered by mr W last updated on 27/Aug/22

4 x = 9 k + 2

\Rightarrow x = 9 m + 5

7 (9 m + 5) = 13 h + 2

13 h - 63 m = 33

\Rightarrow m = 13 n + 10

\Rightarrow x = 9 (13 n + 10) + 5 = 117 n + 95

Answered by CElcedricjunior last updated on 28/Aug/22

{\begin{cases} 4 x \equiv 2 [9] \\ 7 x \equiv 2 [13] \end{cases} => {\begin{cases} 28 x \equiv 14 [9] \\ 14 x \equiv 4 [13] \end{cases} => {\begin{cases} x \equiv 5 [9] \\ x \equiv 4 [13] \end{cases}

=> \exists p; q \in Z / {\begin{cases} x = 9 p + 5 \\ x = 13 q + 4 \end{cases}

o r x = x \Leftrightarrow 9 p - 13 q = - 1

p g c d (9; 13) = 1 e t 1 / 1

a l o r s c e t t e e q u a t i o n a d m e t d e s o l u t i o n

s o i t (- 3; - 2) u n e s o l u t i o n p a r t i c u l i e e

9 p - 13 q = 9 (- 3) - 13 (- 2)

=> 9 (p + 3) = 13 (q + 2) (1)

=> 9 / 13 (q + 2) o r p g c d (9; 13) = 1

=> 9 / q + 2 \Leftrightarrow \exists k \in Z / q + 2 = 9 k (2)

=> q = 9 k - 2

(2) d a n s (1) => 9 (p + 3) = 13 (9 k)

=> p + 3 = 13 k => p = 13 k - 3

d e c e q u i p r e c \overset{`}{e d e {\begin{cases} x = 9 p + 5 \\ x = 13 q + 4 \end{cases}}

d o n c

x = 9 (13 k - 3) + 5 = 117 k - 22

o u

x = 13 (9 k - 2) + 4 = 117 k - 22

S_{Z} = {117 k - 22 / k \in Z}

S_{I N} = {117 k - 22 / k \in] \frac{22}{117}; \to [}

\dots \dots \dots l e c \overset{´}{e l} \overset{`}{e b r e} c e d r i c j u n i o r \dots \dots \dots \dots