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.
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.
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
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.
Ah....thanks, Guest....I misunderstood the problem....no wonder it seemed so easy.....LOL!!!!!
Solution for (n) (smallest positive integer that satisfies the system of congruencies).
n mod 1103 = 1041
n mod 1303 = 859
n mod 2003 = 1095
n≡1041(mod1103)n≡859(mod1303)n≡1095(mod2003)Let m=1103⋅1303⋅2003=2878729627
1103, 1303, and 2003 are coprime numbers (they are actually prime).
n=1041⋅1303⋅2003⋅[(1303⋅2003)φ(1103)−1(mod1103)]⏟=modulo inverse (1303⋅2003)mod1103⏟=(1303⋅2003)1103−1mod1103⏟=(1303⋅2003)1102mod1103⏟=(2609909(mod1103))1102mod1103⏟=(211)1102mod1103⏟=988+859⋅1103⋅2003⋅[(1103⋅2003)φ(1303)−1(mod1303)]⏟=modulo inverse (1103⋅2003)mod1303⏟=(1103⋅2003)1302−1mod1303⏟=(1103⋅2003)1301mod1303⏟=(2209309(mod1303))1301mod1303⏟=(724)1301mod1303⏟=9+1095⋅1103⋅1303⋅[(1103⋅1303)φ(2003)−1(mod2003)]⏟=modulo inverse (1103⋅1303)mod2003⏟=(1103⋅1303)2002−1mod2003⏟=(1103⋅1303)2001mod2003⏟=(1437209(mod2003))2001mod2003⏟=(1058)2001mod2003⏟=195n=1041⋅1303⋅2003⋅[988]+859⋅1103⋅2003⋅[9]+1095⋅1103⋅1303⋅[195]n=2684312285772+17080167879+306880051725n=3008272505376n(modm)=3008272505376(mod2878729627)=45161n=45161+k⋅2878729627k∈Znmin=45161
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