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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
2898440289010
8442892266101
28926612301-1
2662311131-112
2313110-112-13
13101312-1325
10331-1325-88
313025-88289
So our multiplicative inverse is -88 mod 289 ≡ 201
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
821620120101-1
6202013171-14
201171114-14-45
1714134-4549
14342-4549-241
321149-241290
2120-241290-821
So our multiplicative inverse is 290 mod 821 ≡ 290
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
8018920801010
892801191101
8019187301-8
91731181-89
731841-89-44
1811809-44801
So our multiplicative inverse is -44 mod 801 ≡ 757
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 ≡ 633 × 844-1 (mod 289) ≡ 633 × 201 (mod 289) ≡ 73 (mod 289)
x ≡ 18 × 620-1 (mod 821) ≡ 18 × 290 (mod 821) ≡ 294 (mod 821)
x ≡ 549 × 892-1 (mod 801) ≡ 549 × 757 (mod 801) ≡ 675 (mod 801)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 289 × 821 × 801 = 190052469
  2. We calculate the numbers M1 to M3
    M1=M/m1=190052469/289=657621,   M2=M/m2=190052469/821=231489,   M3=M/m3=190052469/801=237269
  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
    2896576210289010
    6576212892275146101
    289146114301-1
    146143131-12
    1433472-12-95
    32112-9597
    2120-9597-289
    So our multiplicative inverse is 97 mod 289 ≡ 97
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    8212314890821010
    231489821281788101
    82178813301-1
    7883323291-124
    332914-124-25
    2947124-25199
    4140-25199-821
    So our multiplicative inverse is 199 mod 821 ≡ 199
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    8012372690801010
    237269801296173101
    801173410901-4
    1731091641-45
    10964145-45-9
    64451195-914
    451927-914-37
    1972514-3788
    7512-3788-125
    522188-125338
    2120-125338-801
    So our multiplicative inverse is 338 mod 801 ≡ 338
    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
    =  (73 × 657621 × 97 +
       294 × 231489 × 199 +
       675 × 237269 × 338)   mod 190052469
    = 113093865 (mod 190052469)


    So our answer is 113093865 (mod 190052469).


Verification

So we found that x ≡ 113093865
If this is correct, then the following statements (i.e. the original equations) are true:
844x (mod 289) ≡ 633 (mod 289)
620x (mod 821) ≡ 18 (mod 821)
892x (mod 801) ≡ 549 (mod 801)

Let's see whether that's indeed the case if we use x ≡ 113093865.