solve { ((x≡3 (mod 5))),((x≡ 1 (mod 7))),((x ≡ 6 (mod 8))) :}


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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
	$solve { ((x≡3 (mod 5))),((x≡ 1 (mod 7))),((x ≡ 6 (mod 8))) :} x=5m+3=7n+1=8p+6 5m=7n−2=7n^′ +5=7n^(′′) +14+5=7n^(′′) +19 5m=8p+3=8p^′ +16+3=8p^′ +19 7n^(′′) =8p^′ =7×8=56 5m=56+19=75 x_1 =75+3=78=7×11+1=8×9+6 x_1 =78 5m=56k+19=56k+15+4=(56k+4)+3×5 (56k+4)≡0(mod 5) (55k+k+4)≡0(mod 5) (k+4)≡0(mod 5) k=5t−4 x=56k+4+3×5+3=56k+22 x=56(5t−4)+22 x=280t−202 t∈Z another way: ... ... x_1 =78 ∵ (5,7,8)=0 ∴ [5,7,8]=5×7×8=280 x=x_1 +n[5,7,8]=78+280n n∈Z$
	$s o l v e {\begin{cases} x \equiv 3 (m o d 5) \\ x \equiv 1 (m o d 7) \\ x \equiv 6 (m o d 8) \end{cases}$ $x = 5 m + 3 = 7 n + 1 = 8 p + 6$ $5 m = 7 n - 2 = 7 n^{^{'}} + 5 = 7 n^{^{″}} + 14 + 5 = 7 n^{^{″}} + 19$ $5 m = 8 p + 3 = 8 p^{^{'}} + 16 + 3 = 8 p^{^{'}} + 19$ $7 n^{^{″}} = 8 p^{^{'}} = 7 \times 8 = 56$ $5 m = 56 + 19 = 75$ $x_{1} = 75 + 3 = 78 = 7 \times 11 + 1 = 8 \times 9 + 6$ $x_{1} = 78$ $5 m = 56 k + 19 = 56 k + 15 + 4 = (56 k + 4) + 3 \times 5$ $(56 k + 4) \equiv 0 (m o d 5)$ $(55 k + k + 4) \equiv 0 (m o d 5)$ $(k + 4) \equiv 0 (m o d 5)$ $k = 5 t - 4$ $x = 56 k + 4 + 3 \times 5 + 3 = 56 k + 22$ $x = 56 (5 t - 4) + 22$ $x = 280 t - 202 t \in Z$ $a n o t h e r w a y :$ $. . . . . .$ $x_{1} = 78$ $∵ (5, 7, 8) = 0$ $∴ [5, 7, 8] = 5 \times 7 \times 8 = 280$ $x = x_{1} + n [5, 7, 8] = 78 + 280 n n \in Z$
	Answered by floor(10²Eta[1]) last updated on 23/Aug/20

	$x \equiv 6 (\mod 8) \Rightarrow x = 8 a + 6$ $8 a + 6 \equiv 3 (\mod 5) \Rightarrow a \equiv 4 (\mod 5) ∴ a = 5 b + 4$ $\Rightarrow x = 8 (5 b + 4) + 6 = 40 b + 38$ $40 b + 38 \equiv 1 (\mod 7) \Rightarrow b \equiv 1 (\mod 7) ∴ b = 7 c + 1$ $\Rightarrow x = 40 (7 c + 1) + 38$ $∴ x = 280 c + 78, c \in Z$ $x \in {. . ., - 482, - 202, 78, 358, . . .}$
	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

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