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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
926489143701-1
4894371521-12
43752821-12-17
52212102-1736
211021-1736-89
10110036-89926
So our multiplicative inverse is -89 mod 926 ≡ 837
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
5696030569010
603569134101
56934162501-16
3425191-1617
25927-1617-50
971217-5067
7231-5067-251
212067-251569
So our multiplicative inverse is -251 mod 569 ≡ 318
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
25711921901-2
11919651-213
19534-213-41
541113-4154
4140-4154-257
So our multiplicative inverse is 54 mod 257 ≡ 54
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 ≡ 23 × 489-1 (mod 926) ≡ 23 × 837 (mod 926) ≡ 731 (mod 926)
x ≡ 569 × 603-1 (mod 569) ≡ 569 × 318 (mod 569) ≡ 0 (mod 569)
x ≡ 95 × 119-1 (mod 257) ≡ 95 × 54 (mod 257) ≡ 247 (mod 257)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 926 × 569 × 257 = 135411758
  2. We calculate the numbers M1 to M3
    M1=M/m1=135411758/926=146233,   M2=M/m2=135411758/569=237982,   M3=M/m3=135411758/257=526894
  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
    9261462330926010
    146233926157851101
    92685117501-1
    8517511261-112
    7526223-112-25
    26231312-2537
    23372-2537-284
    321137-284321
    2120-284321-926
    So our multiplicative inverse is 321 mod 926 ≡ 321
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    5692379820569010
    237982569418140101
    5691404901-4
    14091551-461
    9514-461-65
    541161-65126
    4140-65126-569
    So our multiplicative inverse is 126 mod 569 ≡ 126
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    2575268940257010
    526894257205044101
    2574453701-5
    4437171-56
    37752-56-35
    72316-35111
    2120-35111-257
    So our multiplicative inverse is 111 mod 257 ≡ 111
    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
    =  (731 × 146233 × 321 +
       0 × 237982 × 126 +
       247 × 526894 × 111)   mod 135411758
    = 11339601 (mod 135411758)


    So our answer is 11339601 (mod 135411758).


Verification

So we found that x ≡ 11339601
If this is correct, then the following statements (i.e. the original equations) are true:
489x (mod 926) ≡ 23 (mod 926)
603x (mod 569) ≡ 569 (mod 569)
119x (mod 257) ≡ 95 (mod 257)

Let's see whether that's indeed the case if we use x ≡ 11339601.