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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
1078540107010
8541077105101
1071051201-1
10525211-153
2120-153-107
So our multiplicative inverse is 53 mod 107 ≡ 53
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
6597540659010
754659195101
6599568901-6
9589161-67
896145-67-104
65117-104111
5150-104111-659
So our multiplicative inverse is 111 mod 659 ≡ 111
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
739189317201-3
1891721171-34
17217102-34-43
172814-43348
2120-43348-739
So our multiplicative inverse is 348 mod 739 ≡ 348
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 ≡ 749 × 854-1 (mod 107) ≡ 749 × 53 (mod 107) ≡ 0 (mod 107)
x ≡ 233 × 754-1 (mod 659) ≡ 233 × 111 (mod 659) ≡ 162 (mod 659)
x ≡ 16 × 189-1 (mod 739) ≡ 16 × 348 (mod 739) ≡ 395 (mod 739)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 107 × 659 × 739 = 52109107
  2. We calculate the numbers M1 to M3
    M1=M/m1=52109107/107=487001,   M2=M/m2=52109107/659=79073,   M3=M/m3=52109107/739=70513
  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
    1074870010107010
    487001107455144101
    1074421901-2
    4419261-25
    19631-25-17
    61605-17107
    So our multiplicative inverse is -17 mod 107 ≡ 90
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    659790730659010
    79073659119652101
    6596521701-1
    65279311-194
    7170-194-659
    So our multiplicative inverse is 94 mod 659 ≡ 94
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    739705130739010
    7051373995308101
    739308212301-2
    3081232621-25
    12362161-25-7
    6261115-712
    611610-712-739
    So our multiplicative inverse is 12 mod 739 ≡ 12
    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
    =  (0 × 487001 × 90 +
       162 × 79073 × 94 +
       395 × 70513 × 12)   mod 52109107
    = 27191161 (mod 52109107)


    So our answer is 27191161 (mod 52109107).


Verification

So we found that x ≡ 27191161
If this is correct, then the following statements (i.e. the original equations) are true:
854x (mod 107) ≡ 749 (mod 107)
754x (mod 659) ≡ 233 (mod 659)
189x (mod 739) ≡ 16 (mod 739)

Let's see whether that's indeed the case if we use x ≡ 27191161.