find the value of x such that { ((x=2 (mod 5))),((x=3 (mod 8) )),((x=2 (mod 3))) :}


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Question Number 127111 by benjo_mathlover last updated on 26/Dec/20

$f i n d t h e v a l u e o f x s u c h t h a t$ ${\begin{cases} x = 2 (m o d 5) \\ x = 3 (m o d 8) \\ x = 2 (m o d 3) \end{cases}$
	Answered by liberty last updated on 04/Jan/21

	$g i v e n {\begin{cases} x = 2 (m o d 5) . . . (i) \\ x = 3 (m o d 8) . . . (i i) \\ x = 2 (m o d 3) . . . (i i i) \end{cases}$ $f o r (i) \Rightarrow 24 a \equiv 2 (m o d 5)$ $- a \equiv 2 (m o d 5); a \equiv - 2 (m o d 5)$ $f o r (i i) \Rightarrow 15 b \equiv 3 (m o d 8)$ $- b \equiv 3 (m o d 8); b \equiv - 3 (m o d 8)$ $f o r (i i i) \Rightarrow 40 c \equiv 2 (m o d 3)$ $c \equiv 2 (m o d 3)$ $n o w w e h a v e t h e g e n e r a l s o l u t i o n$ $∴ 24 (- 2) + 15 (- 3) + 40 (2) + 120 k; k \in Z$ $i . e : 120 k - 13; k \in Z or 120 k + 107; k \in Z$
	Answered by physicstutes last updated on 27/Dec/20

	$x \equiv 2 (\mod 3) . . . . . (i) \Rightarrow x = 3 t + 2, t \in Z . . . . . . (i)$ $x \equiv 3 (\mod 8) . . . . . . (i i)$ $(i) in (i) \Rightarrow 3 t + 2 \equiv 3 (\mod 8)$ $3 t \equiv - 1 (\mod 8)$ $but - 1 \equiv 7 (\mod 8) therefore$ $3 t \equiv 7 (\mod 8) by trial and error we find a number t which when$ $multiplied by 3 and divided by 8 gives a remainder of 7 .$ $so i find t = 3 after some tries .$ $\Rightarrow t \equiv 3 (\mod 8) \Rightarrow t = 8 s + 3, s \in Z . . . . . . (i i i)$ $putting (i i i) in (i) \Rightarrow x = 3 (8 s + 3) + 2 = 24 s + 11$ $so : x = 24 s + 11 . . . . . (i v)$ $x \equiv 2 (\mod 5) . . . . . . (v)$ $(i v) in (v) \Rightarrow 24 s + 11 \equiv 2 (\mod 5)$ $24 s \equiv - 9 (\mod 5)$ $but - 9 \equiv 1 (\mod 5) therefore 24 s \equiv 1 (\mod 5) by trial and error$ $s \equiv 4 (\mod 5) \Rightarrow s = 5 u + 4 . . . . . (v i i) u \in Z$ $\Rightarrow x = 24 (5 u + 4) + 11$ $x = 120 u + 107$ $x \equiv 107 (\mod 120)$
	Answered by floor(10²Eta[1]) last updated on 27/Dec/20

	$x \equiv 3 (\mod 8) \Rightarrow x = 8 a + 3, a \in Z$ $8 a + 3 \equiv 3 a + 3 \equiv 2 (\mod 5)$ $3 a \equiv 4 (\mod 5) \Rightarrow a \equiv 3 (\mod 5)$ $\Rightarrow a = 5 b + 3$ $\Rightarrow x = 8 (5 b + 3) + 3 = 40 b + 27, b \in Z$ $40 b + 27 \equiv b \equiv 2 (\mod 3) \Rightarrow b = 3 c + 2$ $\Rightarrow x = 40 (3 c + 2) + 27$ $\Rightarrow x = 120 c + 107, c \in Z$
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