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Question Number 153407 by liberty last updated on 07/Sep/21

	Answered by Rasheed.Sindhi last updated on 08/Sep/21

	$x \equiv 2 (m o d 5) . . . . . . . . (i)$ $x \equiv 1 (m o d 3) . . . . . . . . (i i)$ $x \equiv 6 (m o d 14) . . . . . . (i i i)$ $\underset{\underset{}{-}}{x \equiv 5 \underset{―}{(m o d 11) . . . . . . . (i v)}}$ $(i i i) : 6, 20, 34,$ $(i i i) & (i i) : 34, 76, 118, 160, 202$ $(i) & (i i) & (i i i) : 202, 412$ $(i) & (i i) & (i i i) & (i v) : 412$
	Answered by Rasheed.Sindhi last updated on 13/Sep/21
	$By Chinese Remainder Theorem { ((x≡2(mod 5))),((x≡1(mod 3))),((x≡6(mod 14))),((x≡5(mod 11))) :} a_1 =2,a_2 =1,a_3 =6,a_4 =5 M=m_1 m_2 m_3 m_4 =5×3×14×11=2310 M_1 =(M/m_1 )=((2310)/5)=462 M_2 =(M/m_2 )=((2310)/3)=770 M_3 =(M/m_3 )=((2310)/(14))=165 M_4 =(M/m_4 )=((2310)/(11))=210 { ((462x≡1(mod 5)⇒x_1 =3)),((770x≡1(mod 3)⇒x_2 =2)),((165x≡1(mod 14)⇒x_3 =9)),((210x≡1(mod 11)⇒x_4 =1)) :} x=a_1 M_1 x_1 +a_2 M_2 x_2 +a_3 M_3 x_3 +a_4 M_4 x_4 2.462.3+1.770.2+6.165.9+5.210.1=14272 x=14272≡412(mod 2310)$
	$B y C h i n e s e R e m a i n d e r T h e o r e m$ ${\begin{cases} x \equiv 2 (m o d 5) \\ x \equiv 1 (m o d 3) \\ x \equiv 6 (m o d 14) \\ x \equiv 5 (m o d 11) \end{cases}$ $a_{1} = 2, a_{2} = 1, a_{3} = 6, a_{4} = 5$ $M = m_{1} m_{2} m_{3} m_{4} = 5 \times 3 \times 14 \times 11 = 2310$ $M_{1} = \frac{M}{m_{1}} = \frac{2310}{5} = 462$ $M_{2} = \frac{M}{m_{2}} = \frac{2310}{3} = 770$ $M_{3} = \frac{M}{m_{3}} = \frac{2310}{14} = 165$ $M_{4} = \frac{M}{m_{4}} = \frac{2310}{11} = 210$ ${\begin{cases} 462 x \equiv 1 (m o d 5) \Rightarrow x_{1} = 3 \\ 770 x \equiv 1 (m o d 3) \Rightarrow x_{2} = 2 \\ 165 x \equiv 1 (m o d 14) \Rightarrow x_{3} = 9 \\ 210 x \equiv 1 (m o d 11) \Rightarrow x_{4} = 1 \end{cases}$ $x = a_{1} M_{1} x_{1} + a_{2} M_{2} x_{2} + a_{3} M_{3} x_{3} + a_{4} M_{4} x_{4}$ $2.462.3 + 1.770.2 + 6.165.9 + 5.210.1 = 14272$ $x = 14272 \equiv 412 (m o d 2310)$
	Commented by liberty last updated on 13/Sep/21

	$y e s . . .$
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