## inverse chinese remainder theorem

hi all im new on the forum

I wonder if is possible to find a method that proofs that a number IS NOT a solution of a set of congruences
Maybe using the chinese remainder theorem??

best regards
japam
 Well, simply plugging in the number in the congruences and seeing it they become true should work, right?
 thats a solution but i was thinking in other possibility, a mathematical condition involving the gcd or MCM, for example im sure that it exists but i dont know how to probe that

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## inverse chinese remainder theorem

I can't imagine any way that could be simpler than plugging your number into the equation.

Is there something unusual about your problem that this is not a desirable technique?
 this method is disadvantageous when you have a big number , or a big number of congruences to check And suppose we want to know all the numbers that dont fit anyone of a set of congruences In this case, clearly we need a more analytical method

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 this method is disadvantageous when you have a big number , or a big number of congruences to check
I don't see why.

Anyways, it seems that what you really want to do is to spend a lot of time precomputing information about a system of congruences so that, in the future, you can tell quickly if a number satisfies the system.

The easiest way I can imagine to do this is to simply solve the system of congruences, and then your test is simply comparing a given number to the solution.

e.g. if the system was

3x = 4 (mod 5)
x = 7 (mod 11)

The solution is

x = 18 (mod 55)

So for any given integer n, you simply check if n = 18 (mod 55).
 No, i think is wrong for example X = 2 (mod 3) X = 3 (mod 5) the answer is X = 8 (mod 15) Now i check the num 11 that doesnt fit the last congruence but 11 fits the first so its not a sufficient proof
 Hurkyl, This math is new to me. How do you obtain x = 18 (mod 55)?
 Recognitions: Gold Member Science Advisor Staff Emeritus Ok, so you want integers that satisfy none of the congruences. Again, I cannot divine a better method than simply plugging the integer into each of the congruences. You could achieve some speed-up, possibly, by computing a divisibility test for each of your moduli. Or, better yet, make a giant array whose length is equal to the LCM of all the moduli, and store "true" and "false" for each entry. Loren: The problem is small enough that I did it by trial and error. [:)] In general, you can use the Chinese Remainder theorem.

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 Quote by juan avellaneda hi all im new on the forum I wonder if is possible to find a method that proofs that a number IS NOT a solution of a set of congruences Maybe using the chinese remainder theorem?? best regards japam
According to the Chinese Remainder Theorem if we have moduli X,Y,Z...which have no common factor, then a specific congruent solution is unique up to multiples of XYZ...

Proof: Let us suppose that W == r, Mod X, W==m Mod Y, W == s Mod Z....etc. And some other number T also satisifies the same set of congruences. If W is not T, consider W-T = A. If we look at the set of congruences we find: W-T = A == 0 Mod X, Y, Z...since W and T have the same residues in that system.

So we find that A is divisible by X, Y, Z, etc....and so A ==0 Mod XYZ...

This means A = KXYZ...., where K is an integer. Thus a solution to the Chinese Remainder problem is unique up to the product of the relatively prime moduli.

PS It might be added that uiqueness would not work in the following situation:

X==0 Mod 2, X==2 Mod 8, X==10 Mod 16. Thus X = 10 works in all three cases, but the product of the modulli is 256, while X=26 also works!