Solve the equation: x ≡ 3 (mod 5) x ≡ 4 (mod 7) x ≡ 2 (mod 3)


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Question Number 86791 by TawaTawa1 last updated on 31/Mar/20

$Solve the equation :$ $x \equiv 3 (\mod 5)$ $x \equiv 4 (\mod 7)$ $x \equiv 2 (\mod 3)$
	Answered by mr W last updated on 31/Mar/20

	$x = 5 k + 3$ $x = 7 h + 4$ $x = 3 i + 2$ $5 k + 3 = 7 h + 4$ $5 k - 7 h = 1$ $\Rightarrow k = 7 m - 4 \Rightarrow x = 5 (7 m - 4) + 3 = 35 m - 17$ $\Rightarrow h = 5 m - 3$ $35 m - 17 = 3 i + 2$ $35 m - 3 i = 19$ $\Rightarrow m = 3 n + 2$ $\Rightarrow i = 35 n + 17$ $\Rightarrow x = 35 (3 n + 2) - 17 = 105 n + 53$
	Commented by TawaTawa1 last updated on 31/Mar/20

	$How sir please .$ $Please sir . It is the geimetry question i want you to help me .$ $Thanks for your time sir .$
	Commented by TawaTawa1 last updated on 31/Mar/20

	$I got x = 53 for the congruence sir .$
	Commented by Joel578 last updated on 31/Mar/20

	$Sorry sir, it should be$ $2, 5, 8, 11, 14, 17, . . .$
	Commented by mr W last updated on 31/Mar/20

	$t h a n k s! n o w f i x e d .$
	Answered by Joel578 last updated on 31/Mar/20

	$with Chinese Remainder Theorem$ $a_{1} = 3, m_{1} = 5$ $a_{2} = 4, m_{2} = 7$ $a_{3} = 2, m_{3} = 3$ $\Rightarrow m = (m_{1}) (m_{2}) (m_{3}) = 105$ $\Rightarrow M_{1} = \frac{m}{m_{1}} = 21, M_{2} = \frac{m}{m_{2}} = 15, M_{3} = \frac{m}{m_{3}} = 35$ $Now find invers modulus of M_{k} \mod m_{k}, or in other words,$ $find y_{k} such that M_{k} y_{k} \equiv 1 (\mod m_{k})$ $\Rightarrow y_{1} = 1, y_{2} = 1, y_{3} = 2$ $\Rightarrow y = a_{1} M_{1} y_{1} + a_{2} M_{2} y_{2} + a_{3} M_{3} y_{3}$ $= 63 + 60 + 140 = 263$ $\Rightarrow x \equiv y \mod 105 \Rightarrow x = 53 \to smallest positive value$
	Commented by TawaTawa1 last updated on 31/Mar/20

	$God bless you sir$
	Commented by Ar Brandon last updated on 08/Apr/20
	��
	Answered by Rio Michael last updated on 31/Mar/20

	$another method (algebraic)$ $x \equiv 3 (\mod 5) . . . . . . (i)$ $x \equiv 4 (\mod 7) . . . . . . (ii)$ $x \equiv 2 (\mod 3) . . . . . . . (iii)$ $from (i) x = 5 k + 3, k \in Z . . . . . (i) in (ii)$ $5 k + 3 \equiv 4 (\mod 7)$ $5 k \equiv 1 (\mod 7) by trial and error$ $k \equiv 3 (\mod 7) \Rightarrow k = 7 s + 3, s \in Z \Rightarrow x = 5 (7 s + 3) + 3$ $x = 35 s + 18$ $\Rightarrow 35 s + 18 \equiv 2 (\mod 3)$ $35 s \equiv - 16 (\mod 3)$ $but - 16 \equiv 2 (\mod 3) \Rightarrow 35 s \equiv 2 (\mod 3)$ $so s \equiv 1 (\mod 3) \Rightarrow s = 3 t + 1$ $x = 35 (3 t + 1) + 18 \Rightarrow x = 105 t + 53$ $there for x \equiv 53 (\mod 105)$
	Commented by TawaTawa1 last updated on 31/Mar/20

	$God bless you sir$
	Commented by Rio Michael last updated on 31/Mar/20

	$you too sir$
	Commented by TawaTawa1 last updated on 31/Mar/20

	$I am a girl sir$
	Commented by Rio Michael last updated on 31/Mar/20

	$oh welcome madam$
	Answered by mind is power last updated on 31/Mar/20

	$2 x = 1 (5)$ $2 x = 1 (3)$ $2 x = 1 (7)$ $\Rightarrow 3, 5, 7 ∣ (2 x - 1)$ $d i n c e (3, 5, 7) a r e c o p r i m e s$ $\Rightarrow 3.5.7 ∣ 2 x - 1 \Rightarrow 2 x - 1 = 0 (15.7) = 0 (105)$ $\Rightarrow 2 x - 1 = 0 (105)$ $\Rightarrow 2 x = 1 (105)$ $2 . x . 53 = 53 (105) \Rightarrow x = 53 (105)$
	Commented by TawaTawa1 last updated on 31/Mar/20

	$God bless you sir$
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