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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
6838490683010
8496831166101
68316641901-4
166198141-433
191415-433-37
1452433-37107
5411-37107-144
4140107-144683
So our multiplicative inverse is -144 mod 683 ≡ 539
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
4194410419010
441419122101
4192219101-19
2212201-19419
So our multiplicative inverse is -19 mod 419 ≡ 400
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
4274650427010
465427138101
4273811901-11
389421-1145
9241-1145-191
212045-191427
So our multiplicative inverse is -191 mod 427 ≡ 236
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 ≡ 331 × 849-1 (mod 683) ≡ 331 × 539 (mod 683) ≡ 146 (mod 683)
x ≡ 825 × 441-1 (mod 419) ≡ 825 × 400 (mod 419) ≡ 247 (mod 419)
x ≡ 817 × 465-1 (mod 427) ≡ 817 × 236 (mod 427) ≡ 235 (mod 427)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 683 × 419 × 427 = 122197579
  2. We calculate the numbers M1 to M3
    M1=M/m1=122197579/683=178913,   M2=M/m2=122197579/419=291641,   M3=M/m3=122197579/427=286177
  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
    6831789130683010
    178913683261650101
    68365013301-1
    6503319231-120
    3323110-120-21
    23102320-2162
    10331-2162-207
    313062-207683
    So our multiplicative inverse is -207 mod 683 ≡ 476
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    4192916410419010
    29164141969617101
    41917241101-24
    1711161-2425
    11615-2425-49
    651125-4974
    5150-4974-419
    So our multiplicative inverse is 74 mod 419 ≡ 74
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    4272861770427010
    28617742767087101
    4278747901-4
    8779181-45
    79897-45-49
    87115-4954
    7170-4954-427
    So our multiplicative inverse is 54 mod 427 ≡ 54
    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
    =  (146 × 178913 × 476 +
       247 × 291641 × 74 +
       235 × 286177 × 54)   mod 122197579
    = 11361851 (mod 122197579)


    So our answer is 11361851 (mod 122197579).


Verification

So we found that x ≡ 11361851
If this is correct, then the following statements (i.e. the original equations) are true:
849x (mod 683) ≡ 331 (mod 683)
441x (mod 419) ≡ 825 (mod 419)
465x (mod 427) ≡ 817 (mod 427)

Let's see whether that's indeed the case if we use x ≡ 11361851.