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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.

 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.

 Mar 12, 2017
 #1
avatar+130503 
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

 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+130503 
0

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

 

 

cool cool cool

 Mar 12, 2017
 #4
avatar+2236 
+9

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

n mod 1103 = 1041

n mod 1303 = 859

n mod 2003 = 1095

 

n1041(mod1103)n859(mod1303)n1095(mod2003)Let m=110313032003=2878729627

1103, 1303, and 2003 are coprime numbers (they are actually prime).

n=104113032003[(13032003)φ(1103)1(mod1103)]=modulo inverse (13032003)mod1103=(13032003)11031mod1103=(13032003)1102mod1103=(2609909(mod1103))1102mod1103=(211)1102mod1103=988+85911032003[(11032003)φ(1303)1(mod1303)]=modulo inverse (11032003)mod1303=(11032003)13021mod1303=(11032003)1301mod1303=(2209309(mod1303))1301mod1303=(724)1301mod1303=9+109511031303[(11031303)φ(2003)1(mod2003)]=modulo inverse (11031303)mod2003=(11031303)20021mod2003=(11031303)2001mod2003=(1437209(mod2003))2001mod2003=(1058)2001mod2003=195n=104113032003[988]+85911032003[9]+109511031303[195]n=2684312285772+17080167879+306880051725n=3008272505376n(modm)=3008272505376(mod2878729627)=45161n=45161+k2878729627kZnmin=45161

 

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 Mar 13, 2017

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