solve-x-3-mod-5-x-1-mod-7-x-6-mod-8-

Question Number 109410 by bobhans last updated on 23/Aug/20

s o l v e {\begin{cases} x \equiv 3 (m o d 5) \\ x \equiv 1 (m o d 7) \\ x \equiv 6 (m o d 8) \end{cases}

Commented by bobhans last updated on 24/Aug/20

t h a n k y o u a l l m a s t e r

Answered by 1xx last updated on 23/Aug/20

s o l v e {\begin{cases} x \equiv 3 (m o d 5) \\ x \equiv 1 (m o d 7) \\ x \equiv 6 (m o d 8) \end{cases}

x = 5 m + 3 = 7 n + 1 = 8 p + 6

5 m = 7 n - 2 = 7 n^{^{'}} + 5 = 7 n^{^{″}} + 14 + 5 = 7 n^{^{″}} + 19

5 m = 8 p + 3 = 8 p^{^{'}} + 16 + 3 = 8 p^{^{'}} + 19

7 n^{^{″}} = 8 p^{^{'}} = 7 \times 8 = 56

5 m = 56 + 19 = 75

x_{1} = 75 + 3 = 78 = 7 \times 11 + 1 = 8 \times 9 + 6

x_{1} = 78

5 m = 56 k + 19 = 56 k + 15 + 4 = (56 k + 4) + 3 \times 5

(56 k + 4) \equiv 0 (m o d 5)

(55 k + k + 4) \equiv 0 (m o d 5)

(k + 4) \equiv 0 (m o d 5)

k = 5 t - 4

x = 56 k + 4 + 3 \times 5 + 3 = 56 k + 22

x = 56 (5 t - 4) + 22

x = 280 t - 202 t \in Z

a n o t h e r w a y :

\dots \dots

x_{1} = 78

∵ (5, 7, 8) = 0

∴ [5, 7, 8] = 5 \times 7 \times 8 = 280

x = x_{1} + n [5, 7, 8] = 78 + 280 n n \in Z

Answered by floor(10²Eta[1]) last updated on 23/Aug/20

x \equiv 6 (\mod 8) \Rightarrow x = 8 a + 6

8 a + 6 \equiv 3 (\mod 5) \Rightarrow a \equiv 4 (\mod 5) ∴ a = 5 b + 4

\Rightarrow x = 8 (5 b + 4) + 6 = 40 b + 38

40 b + 38 \equiv 1 (\mod 7) \Rightarrow b \equiv 1 (\mod 7) ∴ b = 7 c + 1

\Rightarrow x = 40 (7 c + 1) + 38

∴ x = 280 c + 78, c \in Z

x \in {\dots, - 482, - 202, 78, 358, \dots}

Commented by 1xx last updated on 23/Aug/20

Y e s . I f o r g o t x < 0 . T h a n k y o u!

Answered by john santu last updated on 24/Aug/20