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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
563147312201-3
1471221251-34
12225422-34-19
2522134-1923
22371-1923-180
313023-180563
So our multiplicative inverse is -180 mod 563 ≡ 383
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
4318230431010
8234311392101
43139213901-1
392391021-111
392191-111-210
212011-210431
So our multiplicative inverse is -210 mod 431 ≡ 221
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1393832501-3
38251131-34
2513112-34-7
1312114-711
121120-711-139
So our multiplicative inverse is 11 mod 139 ≡ 11
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 ≡ 502 × 147-1 (mod 563) ≡ 502 × 383 (mod 563) ≡ 283 (mod 563)
x ≡ 539 × 823-1 (mod 431) ≡ 539 × 221 (mod 431) ≡ 163 (mod 431)
x ≡ 442 × 38-1 (mod 139) ≡ 442 × 11 (mod 139) ≡ 136 (mod 139)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 563 × 431 × 139 = 33728767
  2. We calculate the numbers M1 to M3
    M1=M/m1=33728767/563=59909,   M2=M/m2=33728767/431=78257,   M3=M/m3=33728767/139=242653
  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
    563599090563010
    59909563106231101
    563231210101-2
    2311012291-25
    10129314-25-17
    2914215-1739
    141140-1739-563
    So our multiplicative inverse is 39 mod 563 ≡ 39
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    431782570431010
    78257431181246101
    431246118501-1
    2461851611-12
    1856132-12-7
    6123012-7212
    2120-7212-431
    So our multiplicative inverse is 212 mod 431 ≡ 212
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    1392426530139010
    242653139174598101
    1399814101-1
    98412161-13
    411629-13-7
    169173-710
    9712-710-17
    723110-1761
    2120-1761-139
    So our multiplicative inverse is 61 mod 139 ≡ 61
    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
    =  (283 × 59909 × 39 +
       163 × 78257 × 212 +
       136 × 242653 × 61)   mod 33728767
    = 15639860 (mod 33728767)


    So our answer is 15639860 (mod 33728767).


Verification

So we found that x ≡ 15639860
If this is correct, then the following statements (i.e. the original equations) are true:
147x (mod 563) ≡ 502 (mod 563)
823x (mod 431) ≡ 539 (mod 431)
38x (mod 139) ≡ 442 (mod 139)

Let's see whether that's indeed the case if we use x ≡ 15639860.