Math Question and Answers


Question and Answers Forum

All Questions Topic List
Arithmetic Questions
Previous in All Question Next in All Question
Previous in Arithmetic Next in Arithmetic
Question Number 199621 by Mingma last updated on 06/Nov/23

	Answered by deleteduser1 last updated on 06/Nov/23

	$x + y \overset{3}{\equiv} 2 . . . (i)$ $(6 + 6 + y) - (7 + x) = 5 + y - x \overset{11}{\equiv} 0 \Rightarrow x - y \overset{11}{\equiv} 5 . . . (i i)$ $(i) \land (i i) \Rightarrow (x, y) = (8, 3); (5, 0); (1, 7)$ $60000 + 600 + 3 x + y \overset{7}{\equiv} 0$ $\Rightarrow 3 x + y \overset{7}{\equiv} 6 \Rightarrow (x, y) = (8, 3) \Rightarrow 3 x - 5 y = 9$
	Commented by Mingma last updated on 06/Nov/23
	Nice one!
	Answered by Rasheed.Sindhi last updated on 07/Nov/23
	$676xy ^(−) = { ((22500×3+ 1xy ^(−) )),((9600×7+ 4xy ^(−) )),((6100×11+ 5xy ^(−) )) :} 3 ∣ 676xy ^(−) ⇒3 ∣ 1xy ^(−) 7 ∣ 676xy ^(−) ⇒7 ∣ 4xy ^(−) 11 ∣ 676xy ^(−) ⇒11 ∣ 5xy ^(−) 1xy ^(−) ≡0(mod 3) 4xy ^(−) ≡0(mod 7) 5xy ^(−) ≡0(mod 11) xy ^(−) ≡−100≡2(mod 3) xy ^(−) ≡−400≡6(mod 7) xy ^(−) ≡−500≡6(mod 11) Since 3,7 & 11 are relatively prime So by Chinese theorem: xy ^(−) =83⇒x=8,y=3 3x−5y=3(8)−5(3)=9✓$
	$\overset{―}{676 x y} = {\begin{cases} 22500 \times 3 + \overset{―}{1 x y} \\ 9600 \times 7 + \overset{―}{4 x y} \\ 6100 \times 11 + \overset{―}{5 x y} \end{cases}$ $3 ∣ \overset{―}{676 x y} \Rightarrow 3 ∣ \overset{―}{1 x y}$ $7 ∣ \overset{―}{676 x y} \Rightarrow 7 ∣ \overset{―}{4 x y}$ $11 ∣ \overset{―}{676 x y} \Rightarrow 11 ∣ \overset{―}{5 x y}$ $\overset{―}{1 x y} \equiv 0 (m o d 3)$ $\overset{―}{4 x y} \equiv 0 (m o d 7)$ $\overset{―}{5 x y} \equiv 0 (m o d 11)$ $\overset{―}{x y} \equiv - 100 \equiv 2 (m o d 3)$ $\overset{―}{x y} \equiv - 400 \equiv 6 (m o d 7)$ $\overset{―}{x y} \equiv - 500 \equiv 6 (m o d 11)$ $S i n c e 3, 7 & 11 a r e r e l a t i v e l y p r i m e$ $S o b y C h i n e s e t h e o r e m :$ $\overset{―}{x y} = 83 \Rightarrow x = 8, y = 3$ $3 x - 5 y = 3 (8) - 5 (3) = 9 ✓$
	Answered by Rasheed.Sindhi last updated on 07/Nov/23
	${ (( 676xy ^(−) ≡0(mod 3))),(( 676xy ^(−) ≡0(mod 7))),(( 676xy ^(−) ≡0(mod 11))) :} { (( xy ^(−) ≡−67600(mod 3))),(( xy ^(−) ≡−67600(mod 7))),(( xy ^(−) ≡−67600(mod 11))) :} { (( xy ^(−) ≡−67600+⌈((67600)/3)⌉×3(mod 3))),(( xy ^(−) ≡−67600+⌈((67600)/7)⌉×7(mod 7))),(( xy ^(−) ≡−67600+⌈((67600)/(11))⌉×11(mod 11))) :} { (( xy ^(−) ≡2(mod 3))),(( xy ^(−) ≡6(mod 7))),(( xy ^(−) ≡6(mod 11))) :} Since 3,7 & 11 are pairly coprime So using Chinese theorem: xy ^(−) =83$
	${\begin{cases} \overset{―}{676 x y} \equiv 0 (m o d 3) \\ \overset{―}{676 x y} \equiv 0 (m o d 7) \\ \overset{―}{676 x y} \equiv 0 (m o d 11) \end{cases}$ ${\begin{cases} \overset{―}{x y} \equiv - 67600 (m o d 3) \\ \overset{―}{x y} \equiv - 67600 (m o d 7) \\ \overset{―}{x y} \equiv - 67600 (m o d 11) \end{cases}$ ${\begin{cases} \overset{―}{x y} \equiv - 67600 + ⌈ \frac{67600}{3} ⌉ \times 3 (m o d 3) \\ \overset{―}{x y} \equiv - 67600 + ⌈ \frac{67600}{7} ⌉ \times 7 (m o d 7) \\ \overset{―}{x y} \equiv - 67600 + ⌈ \frac{67600}{11} ⌉ \times 11 (m o d 11) \end{cases}$ ${\begin{cases} \overset{―}{x y} \equiv 2 (m o d 3) \\ \overset{―}{x y} \equiv 6 (m o d 7) \\ \overset{―}{x y} \equiv 6 (m o d 11) \end{cases}$ $S i n c e 3, 7 & 11 a r e p a i r l y c o p r i m e$ $S o u s i n g C h i n e s e t h e o r e m :$ $\overset{―}{x y} = 83$
	Commented by Rasheed.Sindhi last updated on 09/Nov/23
	${ (( z≡2(mod 3))),(( z≡6(mod 7))),(( z≡6(mod 11))) :} ;z= xy ^(−) determinant ((r_i ,m_i ,x_i ^★ ,(r_i m_i x_i )),(2,(77),2,(308)),(6,(33),3,(594)),(6,(21),(10),(1260))) 2162 ^★ 77x_1 ≡1(mod 3)⇒2x_1 ≡1(mod 3)⇒x_1 =2 33x_2 ≡1(mod 7)⇒5x_2 ≡1(mod 7)⇒x_2 =3 21x_3 ≡1(mod 11)⇒10x_3 ≡1(mod 11)⇒x_3 =10 3.7.11=231 z≡2162(mod 231) z≡2162−231(9)(mod 231) z≡83(mod 231)$
	${\begin{cases} z \equiv 2 (m o d 3) \\ z \equiv 6 (m o d 7) \\ z \equiv 6 (m o d 11) \end{cases}; z = \overset{―}{x y}$ $\begin{array}{cccc} r_{i} & m_{i} & x_{i}^{★} & r_{i} m_{i} x_{i} \\ 2 & 77 & 2 & 308 \\ 6 & 33 & 3 & 594 \\ 6 & 21 & 10 & 1260 \end{array}$ $2162$ $^{★} 77 x_{1} \equiv 1 (m o d 3) \Rightarrow 2 x_{1} \equiv 1 (m o d 3) \Rightarrow x_{1} = 2$ $33 x_{2} \equiv 1 (m o d 7) \Rightarrow 5 x_{2} \equiv 1 (m o d 7) \Rightarrow x_{2} = 3$ $21 x_{3} \equiv 1 (m o d 11) \Rightarrow 10 x_{3} \equiv 1 (m o d 11) \Rightarrow x_{3} = 10$ $3.7.11 = 231$ $z \equiv 2162 (m o d 231)$ $z \equiv 2162 - 231 (9) (m o d 231)$ $z \equiv 83 (m o d 231)$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum