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Welcome to ChineseRemainderTheorem.com!

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Use the calculator below to get a step-by-step calculation of the Chinese Remainder Theory. Just enter the numbers you would like and press "Calculate" .


This removes all numbers from the textboxes, such that you can fill in your own.

This fills all textboxes with random numbers. If you fill in random numbers yourself, it is very likely that those numbers do not have a solution. To avoid disappointment, use this button instead! It only uses random numbers that do have a solution.

This button is similar to the "Clear everything" button, but only clears the left column.
This is useful if you want your equations to be of the form x ≡ a (mod m) rather than bx ≡ a (mod m).
In that case, it can be especially useful after using the random numbers button.

Do you want to use more equations? Go ahead and use this button. It adds another row that you can fill in. Not sure what numbers to put in this newly added row? Use the random numbers button again!

Do you have too many rows? Use one of these buttons to remove a row. You can always add a row again using the yellow "Add a row" button.

Are you ready to view a full step-by-step Chinese Remainder Theorem calculation for the numbers you have entered? Then use this button!

Want to know more?


Transform the equations

You used one or more of the fields on the left, so your equations are of the form bx ≡ a mod m.
We want them to be of the form x ≡ a mod m, so we need to move the values on the left to the right side of the equation.
For a more detailed explanation about how this works, see this part of our page about how to execute the Chinese Remainder algorithm.

First, we calculate the inverses of the leftmost value on each row:

nbqr t1t2t3
887482140501-1
4824051771-12
40577520-12-11
77203172-1135
201713-1135-46
1735235-46265
3211-46265-311
2120265-311887
So our multiplicative inverse is -311 mod 887 ≡ 576
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
6379550637010
9556371318101
6373182101-2
318131801-2637
So our multiplicative inverse is -2 mod 637 ≡ 635
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
5736760573010
6765731103101
57310355801-5
103581451-56
5845113-56-11
4513366-1139
13621-1139-89
616039-89573
So our multiplicative inverse is -89 mod 573 ≡ 484
Source: ExtendedEuclideanAlgorithm.com

Click on any row to reveal a more detailed calculation of each multiplicative inverse.

Now that we now the inverses, let's move the leftmost value on each row to the right of the equation:

x ≡ 429 × 482-1 (mod 887) ≡ 429 × 576 (mod 887) ≡ 518 (mod 887)
x ≡ 453 × 955-1 (mod 637) ≡ 453 × 635 (mod 637) ≡ 368 (mod 637)
x ≡ 732 × 676-1 (mod 573) ≡ 732 × 484 (mod 573) ≡ 174 (mod 573)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 887 × 637 × 573 = 323755887
  2. We calculate the numbers M1 to M3
    M1=M/m1=323755887/887=365001,   M2=M/m2=323755887/637=508251,   M3=M/m3=323755887/573=565019
  3. We now calculate the modular multiplicative inverses M1-1 to M3-1
    Have a look at the page that explains how to calculate modular multiplicative inverse.
    Using, for example, the Extended Euclidean Algorithm, we will find that:

    nbqr t1t2t3
    8873650010887010
    365001887411444101
    887444144301-1
    444443111-12
    44314430-12-887
    So our multiplicative inverse is 2 mod 887 ≡ 2
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    6375082510637010
    508251637797562101
    63756217501-1
    562757371-18
    753721-18-17
    3713708-17637
    So our multiplicative inverse is -17 mod 637 ≡ 620
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    5735650190573010
    56501957398641101
    57341134001-13
    4140111-1314
    401400-1314-573
    So our multiplicative inverse is 14 mod 573 ≡ 14
    Source: ExtendedEuclideanAlgorithm.com
  4. Now we can calculate x with the equation we saw earlier
    x = (a1 × M1 × M1-1   +   a2 × M2 × M2-1   + ... +   ak × Mk × Mk-1)   mod M
    =  (518 × 365001 × 2 +
       368 × 508251 × 620 +
       174 × 565019 × 14)   mod 323755887
    = 193688499 (mod 323755887)


    So our answer is 193688499 (mod 323755887).


Verification

So we found that x ≡ 193688499
If this is correct, then the following statements (i.e. the original equations) are true:
482x (mod 887) ≡ 429 (mod 887)
955x (mod 637) ≡ 453 (mod 637)
676x (mod 573) ≡ 732 (mod 573)

Let's see whether that's indeed the case if we use x ≡ 193688499.