Chinese remainder theorem

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Information reference: https://zh.wikipedia.org/wiki/%E4%B8%AD%E5%9B%BD%E5%89%A9%E4%BD%99%E5%AE%9A%E7%90%86

"There is no known number, 33 of the remaining two, 55 of the remaining three, 77 of the remaining two." Asking for geometry? "This is the source of the Chinese remainder theorem, what does it mean?" That is, if the number of items, if three three number, there will be two, if five five numbers, there will be three, if it is 77, then there will be two, then how many of them? (Source wiki)

I would like to discuss this issue from a special to a general point;

The question can actually be abstracted to the mathematics problem as follows: Now we think, how can we find such a number x?

First we define A[1 2 3] = 2, 3, 2;m[1 2 3] = 3, 5, 7; We think, if there is such an x, then what is the nature of it? First it to meet the Mi modulus when the corresponding value of the AI, good then we from the programming cycle to think, whether there is a ti for ti% mi = ai, while ti% MJ (j! = i) = 0 if present, then we x can be equal to T1 + T2 + T3 (because In order to assign the module to TI separately, there are two values of 0) If so, how to construct? Well, at least we have a clue now. Then ti how to build, must first have an AI% mi = ai, then we can find a ti/ai, it satisfies ti/ai% mi = 1, at the same time by ti% MJ (j! = i) = 0, ti/ai% mi = 1, Ti% mi = ai can get this T I/ai This item has a property of ti/ai% MJ = 0; Okay, now about

Ti/ai This item has a certain nature: Ti/ai% MJ =0, ti/ai% mi = 1, so through the structure of the TI/AI is a multiple of MJ, at the same time, it will be in MI under two reciprocal of the product, so the use of MI for the multiplication of MI, but excluding item I, while making Ti = mi mode m I's Number theory reciprocal, also that satisfies ti * Mi = 1 (mod mi), Ti/ai = Ti * MI;

(here to add a description, what is the number of modulo mi countdown, that is, when the MI is determined, you can use the TI * mi% mi = = 1 of the figure, this can be cycled to seek, the maximum number of cycles is Mi, because Ti < mi, hit a minimal ti on it can be)

That's now ok ti = ai * ti * Mi, while for (S) we have a ti * mi% mi = 1, which means

So you can determine T1, T2, T3, you can also find X.

For the general proof of the principle, I think the proof on the wiki is very good.

There is one to add, why x1-x2 ==0 (mod mi) and 22 mi between coprime can be released between the two solutions of the number of related m-multiples? is because x1-x2 = = 0 (mod mi) is indicative of x1-x2

At least there is a total of MI as an approximate (the above equation is set up for all MI), and M is the product of all MI, so you can get x1-x2 is a multiple of M, x1-x2 <= m (assuming x1 > x2)

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Chinese remainder theorem