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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
5298220529010
8225291293101
529293123601-1
2932361571-12
2365748-12-9
578712-965
8180-965-529
So our multiplicative inverse is 65 mod 529 ≡ 65
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
67974067010
974671436101
673613101-1
3631151-12
31561-12-13
51502-1367
So our multiplicative inverse is -13 mod 67 ≡ 54
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
4788610478010
8614781383101
47838319501-1
38395431-15
953312-15-156
32115-156161
2120-156161-478
So our multiplicative inverse is 161 mod 478 ≡ 161
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 ≡ 103 × 822-1 (mod 529) ≡ 103 × 65 (mod 529) ≡ 347 (mod 529)
x ≡ 869 × 974-1 (mod 67) ≡ 869 × 54 (mod 67) ≡ 26 (mod 67)
x ≡ 391 × 861-1 (mod 478) ≡ 391 × 161 (mod 478) ≡ 333 (mod 478)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 529 × 67 × 478 = 16941754
  2. We calculate the numbers M1 to M3
    M1=M/m1=16941754/529=32026,   M2=M/m2=16941754/67=252862,   M3=M/m3=16941754/478=35443
  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
    529320260529010
    3202652960286101
    529286124301-1
    2862431431-12
    24343528-12-11
    43281152-1113
    2815113-1113-24
    15131213-2437
    13261-2437-246
    212037-246529
    So our multiplicative inverse is -246 mod 529 ≡ 283
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    67252862067010
    2528626737744101
    67416301-16
    43111-1617
    3130-1617-67
    So our multiplicative inverse is 17 mod 67 ≡ 17
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    478354430478010
    354434787471101
    4787165201-6
    71521191-67
    5219214-67-20
    1914157-2027
    14524-2027-74
    541127-74101
    4140-74101-478
    So our multiplicative inverse is 101 mod 478 ≡ 101
    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
    =  (347 × 32026 × 283 +
       26 × 252862 × 17 +
       333 × 35443 × 101)   mod 16941754
    = 10065101 (mod 16941754)


    So our answer is 10065101 (mod 16941754).


Verification

So we found that x ≡ 10065101
If this is correct, then the following statements (i.e. the original equations) are true:
822x (mod 529) ≡ 103 (mod 529)
974x (mod 67) ≡ 869 (mod 67)
861x (mod 478) ≡ 391 (mod 478)

Let's see whether that's indeed the case if we use x ≡ 10065101.