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Find the smallest positive number that will satisfy the following modular equations:
n mod 1,103 = 1,041, n mod 1,303 = 859, n mod 2,003 = 1,095.
Thanks for any help.

Guest Mar 12, 2017

Best Answer 

 #2
avatar
+5

Very nice going CPhill!!!. But, with little help from a simple computer code, we can do a bit better!.

 

A*1,103 + 1,041 =B*1,303 + 859 = C*2,003 + 1,095, solve for A, B, C.

 

A = 40, B = 34, C = 22. Therefore, the smallest positive number will be: 40 x 1,103 + 1,041 =45,161.
The general formula will then be:


LCM{1,103, 1,303, 2003} = 2,878,729,627

2,878,729,627D + 45,161 = n. For D =0, 1, 2.....etc. we have:


45,161, 2,878,774,788, 5,757,504,415.....etc.

Guest Mar 12, 2017
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4+0 Answers

 #1
avatar+75302 
0

These are all trivial answers......!!!!

 

n mod 1,103 = 1,041     n = 1041

n mod 1,303 = 859        n = 859

n mod 2,003 = 1,095     n = 1095

 

 

cool cool cool

CPhill  Mar 12, 2017
 #2
avatar
+5
Best Answer

Very nice going CPhill!!!. But, with little help from a simple computer code, we can do a bit better!.

 

A*1,103 + 1,041 =B*1,303 + 859 = C*2,003 + 1,095, solve for A, B, C.

 

A = 40, B = 34, C = 22. Therefore, the smallest positive number will be: 40 x 1,103 + 1,041 =45,161.
The general formula will then be:


LCM{1,103, 1,303, 2003} = 2,878,729,627

2,878,729,627D + 45,161 = n. For D =0, 1, 2.....etc. we have:


45,161, 2,878,774,788, 5,757,504,415.....etc.

Guest Mar 12, 2017
 #3
avatar+75302 
0

Ah....thanks, Guest....I misunderstood the problem....no wonder it seemed so easy.....LOL!!!!!

 

 

cool cool cool

CPhill  Mar 12, 2017
 #4
avatar+644 
+5

Solution for (n) (smallest positive integer that satisfies the system of congruencies).

n mod 1103 = 1041

n mod 1303 = 859

n mod 2003 = 1095

 

\(\begin{array}{rcll} n &\equiv& {\color{red}1041} \pmod {{\color{green}1103}} \\ n &\equiv& {\color{red}859} \pmod {{\color{green}1303}} \\n &\equiv& {\color{red}1095} \pmod {{\color{green}2003}} \\ \text{Let } m &=&1103 \cdot 1303\cdot 2003 = 2878729627 \\ \end{array} \)

\(\text {1103, 1303, and 2003 are coprime numbers (they are actually prime).}\\\)

\(\small{ \begin{array}{l} n = {\color{red}1041} \cdot {\color{green}1303\cdot 2003} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ { (\color{green}1303\cdot 2003)}^{\varphi({\color{green}1103}) -1 } \pmod {{\color{green}1103}} ] }_{=\text{modulo inverse }(1303\cdot 2003) mod 1103 }}_{=(1303\cdot 2003)^{1103-1} \mod {1103}} }_{=(1303\cdot 2003)^{1102} \mod {1103}} }_{=(2609909\pmod{1103})^{1102} \mod {1103}} }_{=(211)^{1102} \mod {1103}} }_{=988} + {\color{red}859} \cdot {\color{green}1103\cdot 2003} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ { (\color{green}1103\cdot 2003) }^{\varphi({\color{green}1303}) -1} \pmod {{\color{green}1303}} ] }_{=\text{modulo inverse } (1103\cdot 2003) mod 1303 } }_{=(1103\cdot 2003)^{1302-1} \mod {1303}} }_{=(1103\cdot 2003)^{1301} \mod {1303}} }_{=(2209309\pmod{1303})^{1301} \mod {1303}} }_{=(724)^{1301} \mod {1303}} }_{=9} +{\color{red}{1095}} \cdot {\color{green}1103\cdot 1303} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ { (\color{green}1103\cdot 1303) }^{\varphi({\color{green}2003}) -1 } \pmod {{\color{green}2003}} ] }_{=\text{modulo inverse } (1103\cdot 1303) mod 2003 } }_{=(1103\cdot 1303)^{2002-1} \mod {2003}} }_{=(1103\cdot 1303)^{2001} \mod {2003}} }_{=(1437209\pmod{2003})^{2001} \mod {2003}} }_{=(1058)^{2001} \mod {2003}} }_{=195}\\\\ n = {\color{red}{1041}} \cdot {\color{green}{1303}\cdot 2003} \cdot [988] + {\color{red}859} \cdot {\color{green}1103\cdot 2003} \cdot [9] + {\color{red}1095} \cdot {\color{green}1103\cdot 1303} \cdot [195] \\ n = 2684312285772 + 17080167879 + 306880051725 \\ n = 3008272505376 \\\\ n \pmod {m}\\ = 3008272505376 \pmod {2878729627} \\ = 45161 \\\\ n = 45161 + k\cdot 2878729627 \qquad k \in Z\\\\ \mathbf{n_{min}} \mathbf{=} \mathbf{45161} \end{array}} \)

 

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