# Inverse chinese remainder theorem

1. Feb 27, 2004

### 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

2. Feb 27, 2004

### Muzza

Well, simply plugging in the number in the congruences and seeing it they become true should work, right?

3. Feb 27, 2004

### juan avellaneda

method

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

4. Feb 27, 2004

### Hurkyl

Staff Emeritus
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?

5. Feb 27, 2004

### juan avellaneda

method

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

6. Feb 27, 2004

### Hurkyl

Staff Emeritus
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).

7. Feb 27, 2004

### juan avellaneda

counterexample

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

8. Feb 27, 2004

### Loren Booda

Hurkyl,

This math is new to me. How do you obtain x = 18 (mod 55)?

9. Feb 27, 2004

### Hurkyl

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.

10. Jul 6, 2004

### robert Ihnot

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!

Last edited: Jul 6, 2004