Is There a Solution for Modulus with 3 Equations?

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coolul007
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I am wondering if there is a way to solve the following:

n == a mod(m)

m == b mod(r)

n == x mod(r), I need to know x, without dealing with a gigantic number(n),

I can find a and b, as I chose m to give me predictable results.

I appreciate any insight...
 
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Sounds like a Chinese Remainder Theorem problem. Check out the concept online or in any book on number theory.
 
It doesn' seem to be Chinese Remainder...I've looked into that. If it was Chinese Remainder it would be solving for a number with known mods, I know the number but need the missing mod.

x == a mod(m)
x == b mod(r)
 
Ahhh, I see. Well, never mind then!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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