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Question-199621

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Question Number 199621 by Mingma last updated on 06/Nov/23
Answered by AST last updated on 06/Nov/23
x+y≡^3 2...(i) (6+6+y)−(7+x)=5+y−x≡^(11) 0⇒x−y≡^(11) 5...(ii) (i)∧(ii)⇒(x,y)=(8,3);(5,0);(1,7) 60000+600+3x+y≡^7 0 ⇒3x+y≡^7 6⇒(x,y)=(8,3)⇒3x−5y=9
$${x}+{y}\overset{\mathrm{3}} {\equiv}\mathrm{2}…\left({i}\right) \\ $$$$\left(\mathrm{6}+\mathrm{6}+{y}\right)−\left(\mathrm{7}+{x}\right)=\mathrm{5}+{y}−{x}\overset{\mathrm{11}} {\equiv}\mathrm{0}\Rightarrow{x}−{y}\overset{\mathrm{11}} {\equiv}\mathrm{5}…\left({ii}\right) \\ $$$$\left({i}\right)\wedge\left({ii}\right)\Rightarrow\left({x},{y}\right)=\left(\mathrm{8},\mathrm{3}\right);\left(\mathrm{5},\mathrm{0}\right);\left(\mathrm{1},\mathrm{7}\right) \\ $$$$\mathrm{60000}+\mathrm{600}+\mathrm{3}{x}+{y}\overset{\mathrm{7}} {\equiv}\mathrm{0} \\ $$$$\Rightarrow\mathrm{3}{x}+{y}\overset{\mathrm{7}} {\equiv}\mathrm{6}\Rightarrow\left({x},{y}\right)=\left(\mathrm{8},\mathrm{3}\right)\Rightarrow\mathrm{3}{x}−\mathrm{5}{y}=\mathrm{9} \\ $$
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✓
$$\overline {\:\mathrm{676}{xy}\:}=\begin{cases}{\mathrm{22500}×\mathrm{3}+\overline {\:\mathrm{1}{xy}\:}\:}\\{\mathrm{9600}×\mathrm{7}+\overline {\:\mathrm{4}{xy}\:}}\\{\mathrm{6100}×\mathrm{11}+\overline {\:\mathrm{5}{xy}\:}}\end{cases} \\ $$$$\mathrm{3}\:\mid\:\overline {\:\mathrm{676}{xy}\:}\Rightarrow\mathrm{3}\:\mid\:\overline {\:\mathrm{1}{xy}\:} \\ $$$$\mathrm{7}\:\mid\:\overline {\:\mathrm{676}{xy}\:}\Rightarrow\mathrm{7}\:\mid\:\overline {\:\mathrm{4}{xy}\:} \\ $$$$\mathrm{11}\:\mid\:\overline {\:\mathrm{676}{xy}\:}\Rightarrow\mathrm{11}\:\mid\:\overline {\:\mathrm{5}{xy}\:} \\ $$$$\overline {\:\mathrm{1}{xy}\:}\equiv\mathrm{0}\left({mod}\:\mathrm{3}\right) \\ $$$$\overline {\:\mathrm{4}{xy}\:}\equiv\mathrm{0}\left({mod}\:\mathrm{7}\right) \\ $$$$\overline {\:\mathrm{5}{xy}\:}\equiv\mathrm{0}\left({mod}\:\mathrm{11}\right) \\ $$$$\:\overline {\:{xy}\:}\equiv−\mathrm{100}\equiv\mathrm{2}\left({mod}\:\mathrm{3}\right) \\ $$$$\:\overline {\:{xy}\:}\equiv−\mathrm{400}\equiv\mathrm{6}\left({mod}\:\mathrm{7}\right) \\ $$$$\:\overline {\:{xy}\:}\equiv−\mathrm{500}\equiv\mathrm{6}\left({mod}\:\mathrm{11}\right) \\ $$$${Since}\:\mathrm{3},\mathrm{7}\:\&\:\mathrm{11}\:{are}\:{relatively}\:{prime} \\ $$$${So}\:{by}\:{Chinese}\:{theorem}: \\ $$$$\:\:\overline {\:{xy}\:}=\mathrm{83}\Rightarrow{x}=\mathrm{8},{y}=\mathrm{3} \\ $$$$\mathrm{3}{x}−\mathrm{5}{y}=\mathrm{3}\left(\mathrm{8}\right)−\mathrm{5}\left(\mathrm{3}\right)=\mathrm{9}\checkmark \\ $$
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
$$\begin{cases}{\overline {\:\mathrm{676}{xy}\:}\equiv\mathrm{0}\left({mod}\:\mathrm{3}\right)}\\{\overline {\:\mathrm{676}{xy}\:}\equiv\mathrm{0}\left({mod}\:\mathrm{7}\right)}\\{\overline {\:\mathrm{676}{xy}\:}\equiv\mathrm{0}\left({mod}\:\mathrm{11}\right)}\end{cases}\: \\ $$$$\begin{cases}{\overline {\:{xy}\:}\equiv−\mathrm{67600}\left({mod}\:\mathrm{3}\right)}\\{\overline {\:{xy}\:}\equiv−\mathrm{67600}\left({mod}\:\mathrm{7}\right)}\\{\overline {\:{xy}\:}\equiv−\mathrm{67600}\left({mod}\:\mathrm{11}\right)}\end{cases}\:\: \\ $$$$\begin{cases}{\overline {\:{xy}\:}\equiv−\mathrm{67600}+\lceil\frac{\mathrm{67600}}{\mathrm{3}}\rceil×\mathrm{3}\left({mod}\:\mathrm{3}\right)}\\{\overline {\:{xy}\:}\equiv−\mathrm{67600}+\lceil\frac{\mathrm{67600}}{\mathrm{7}}\rceil×\mathrm{7}\left({mod}\:\mathrm{7}\right)}\\{\overline {\:{xy}\:}\equiv−\mathrm{67600}+\lceil\frac{\mathrm{67600}}{\mathrm{11}}\rceil×\mathrm{11}\left({mod}\:\mathrm{11}\right)}\end{cases}\:\: \\ $$$$\begin{cases}{\overline {\:{xy}\:}\equiv\mathrm{2}\left({mod}\:\mathrm{3}\right)}\\{\overline {\:{xy}\:}\equiv\mathrm{6}\left({mod}\:\mathrm{7}\right)}\\{\overline {\:{xy}\:}\equiv\mathrm{6}\left({mod}\:\mathrm{11}\right)}\end{cases}\: \\ $$$${Since}\:\mathrm{3},\mathrm{7}\:\&\:\mathrm{11}\:{are}\:{pairly}\:\:{coprime} \\ $$$${So}\:{using}\:{Chinese}\:{theorem}: \\ $$$$\overline {\:{xy}\:}=\mathrm{83} \\ $$
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)
$$\begin{cases}{\:\:{z}\equiv\mathrm{2}\left({mod}\:\mathrm{3}\right)}\\{\:\:{z}\equiv\mathrm{6}\left({mod}\:\mathrm{7}\right)}\\{\:\:{z}\equiv\mathrm{6}\left({mod}\:\mathrm{11}\right)}\end{cases}\:;{z}=\overline {\:{xy}\:}\: \\ $$$$\begin{array}{|c|c|c|c|}{{r}_{{i}} }&\hline{{m}_{{i}} }&\hline{{x}_{{i}} ^{\bigstar} }&\hline{{r}_{{i}} {m}_{{i}} {x}_{{i}} }\\{\mathrm{2}}&\hline{\mathrm{77}}&\hline{\mathrm{2}}&\hline{\mathrm{308}}\\{\mathrm{6}}&\hline{\mathrm{33}}&\hline{\mathrm{3}}&\hline{\mathrm{594}}\\{\mathrm{6}}&\hline{\mathrm{21}}&\hline{\mathrm{10}}&\hline{\mathrm{1260}}\\\hline\end{array} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2162} \\ $$$$\:^{\bigstar} \mathrm{77}{x}_{\mathrm{1}} \equiv\mathrm{1}\left({mod}\:\mathrm{3}\right)\Rightarrow\mathrm{2}{x}_{\mathrm{1}} \equiv\mathrm{1}\left({mod}\:\mathrm{3}\right)\Rightarrow{x}_{\mathrm{1}} =\mathrm{2} \\ $$$$\:\:\:\:\mathrm{33}{x}_{\mathrm{2}} \equiv\mathrm{1}\left({mod}\:\mathrm{7}\right)\Rightarrow\mathrm{5}{x}_{\mathrm{2}} \equiv\mathrm{1}\left({mod}\:\mathrm{7}\right)\Rightarrow{x}_{\mathrm{2}} =\mathrm{3} \\ $$$$\:\:\:\:\mathrm{21}{x}_{\mathrm{3}} \equiv\mathrm{1}\left({mod}\:\mathrm{11}\right)\Rightarrow\mathrm{10}{x}_{\mathrm{3}} \equiv\mathrm{1}\left({mod}\:\mathrm{11}\right)\Rightarrow{x}_{\mathrm{3}} =\mathrm{10} \\ $$$$\: \\ $$$$\:\mathrm{3}.\mathrm{7}.\mathrm{11}=\mathrm{231} \\ $$$${z}\equiv\mathrm{2162}\left({mod}\:\mathrm{231}\right)\: \\ $$$${z}\equiv\mathrm{2162}−\mathrm{231}\left(\mathrm{9}\right)\left({mod}\:\mathrm{231}\right) \\ $$$${z}\equiv\mathrm{83}\left({mod}\:\mathrm{231}\right) \\ $$

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