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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
74736821101-2
368113351-267
11521-267-136
515067-136747
So our multiplicative inverse is -136 mod 747 ≡ 611
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1073110107010
311107297101
1079711001-1
9710971-110
10713-110-11
732110-1132
3130-1132-107
So our multiplicative inverse is 32 mod 107 ≡ 32
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1637260163010
726163474101
1637421501-2
74154141-29
151411-29-11
1411409-11163
So our multiplicative inverse is -11 mod 163 ≡ 152
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 ≡ 680 × 368-1 (mod 747) ≡ 680 × 611 (mod 747) ≡ 148 (mod 747)
x ≡ 235 × 311-1 (mod 107) ≡ 235 × 32 (mod 107) ≡ 30 (mod 107)
x ≡ 958 × 726-1 (mod 163) ≡ 958 × 152 (mod 163) ≡ 57 (mod 163)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 747 × 107 × 163 = 13028427
  2. We calculate the numbers M1 to M3
    M1=M/m1=13028427/747=17441,   M2=M/m2=13028427/107=121761,   M3=M/m3=13028427/163=79929
  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
    747174410747010
    1744174723260101
    747260222701-2
    2602271331-23
    22733629-23-20
    3329143-2023
    29471-2023-181
    414023-181747
    So our multiplicative inverse is -181 mod 747 ≡ 566
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    1071217610107010
    1217611071137102101
    1071021501-1
    10252021-121
    5221-121-43
    212021-43107
    So our multiplicative inverse is -43 mod 107 ≡ 64
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    163799290163010
    7992916349059101
    1635924501-2
    59451141-23
    451433-23-11
    143423-1147
    3211-1147-58
    212047-58163
    So our multiplicative inverse is -58 mod 163 ≡ 105
    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
    =  (148 × 17441 × 566 +
       30 × 121761 × 64 +
       57 × 79929 × 105)   mod 13028427
    = 10434991 (mod 13028427)


    So our answer is 10434991 (mod 13028427).


Verification

So we found that x ≡ 10434991
If this is correct, then the following statements (i.e. the original equations) are true:
368x (mod 747) ≡ 680 (mod 747)
311x (mod 107) ≡ 235 (mod 107)
726x (mod 163) ≡ 958 (mod 163)

Let's see whether that's indeed the case if we use x ≡ 10434991.