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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
24416318101-1
16381211-13
811810-13-244
So our multiplicative inverse is 3 mod 244 ≡ 3
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
83677083010
67783813101
83136501-6
135231-613
5312-613-19
321113-1932
2120-1932-83
So our multiplicative inverse is 32 mod 83 ≡ 32
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
557424113301-1
4241333251-14
1332558-14-21
258314-2167
8180-2167-557
So our multiplicative inverse is 67 mod 557 ≡ 67
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 ≡ 734 × 163-1 (mod 244) ≡ 734 × 3 (mod 244) ≡ 6 (mod 244)
x ≡ 489 × 677-1 (mod 83) ≡ 489 × 32 (mod 83) ≡ 44 (mod 83)
x ≡ 797 × 424-1 (mod 557) ≡ 797 × 67 (mod 557) ≡ 484 (mod 557)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 244 × 83 × 557 = 11280364
  2. We calculate the numbers M1 to M3
    M1=M/m1=11280364/244=46231,   M2=M/m2=11280364/83=135908,   M3=M/m3=11280364/557=20252
  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
    244462310244010
    46231244189115101
    24411521401-2
    11514831-217
    14342-217-70
    321117-7087
    2120-7087-244
    So our multiplicative inverse is 87 mod 244 ≡ 87
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    83135908083010
    13590883163737101
    83372901-2
    379411-29
    9190-29-83
    So our multiplicative inverse is 9 mod 83 ≡ 9
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    557202520557010
    2025255736200101
    557200215701-2
    2001571431-23
    15743328-23-11
    43281153-1114
    2815113-1114-25
    15131214-2539
    13261-2539-259
    212039-259557
    So our multiplicative inverse is -259 mod 557 ≡ 298
    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
    =  (6 × 46231 × 87 +
       44 × 135908 × 9 +
       484 × 20252 × 298)   mod 11280364
    = 9642154 (mod 11280364)


    So our answer is 9642154 (mod 11280364).


Verification

So we found that x ≡ 9642154
If this is correct, then the following statements (i.e. the original equations) are true:
163x (mod 244) ≡ 734 (mod 244)
677x (mod 83) ≡ 489 (mod 83)
424x (mod 557) ≡ 797 (mod 557)

Let's see whether that's indeed the case if we use x ≡ 9642154.