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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
2375410237010
541237267101
2376733601-3
67361311-34
363115-34-7
315614-746
5150-746-237
So our multiplicative inverse is 46 mod 237 ≡ 46
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
4515410451010
541451190101
451905101-5
9019001-5451
So our multiplicative inverse is -5 mod 451 ≡ 446
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
5779210577010
9215771344101
577344123301-1
34423311111-12
233111211-12-5
111111012-552
111110-552-577
So our multiplicative inverse is 52 mod 577 ≡ 52
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 ≡ 637 × 541-1 (mod 237) ≡ 637 × 46 (mod 237) ≡ 151 (mod 237)
x ≡ 694 × 541-1 (mod 451) ≡ 694 × 446 (mod 451) ≡ 138 (mod 451)
x ≡ 567 × 921-1 (mod 577) ≡ 567 × 52 (mod 577) ≡ 57 (mod 577)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 237 × 451 × 577 = 61673799
  2. We calculate the numbers M1 to M3
    M1=M/m1=61673799/237=260227,   M2=M/m2=61673799/451=136749,   M3=M/m3=61673799/577=106887
  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
    2372602270237010
    26022723710981101
    2371237001-237
    So our multiplicative inverse is 1 mod 237 ≡ 1
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    4511367490451010
    13674945130396101
    4519646701-4
    96671291-45
    672929-45-14
    299325-1447
    9241-1447-202
    212047-202451
    So our multiplicative inverse is -202 mod 451 ≡ 249
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    5771068870577010
    106887577185142101
    5771424901-4
    14291571-461
    9712-461-65
    723161-65256
    2120-65256-577
    So our multiplicative inverse is 256 mod 577 ≡ 256
    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
    =  (151 × 260227 × 1 +
       138 × 136749 × 249 +
       57 × 106887 × 256)   mod 61673799
    = 7231021 (mod 61673799)


    So our answer is 7231021 (mod 61673799).


Verification

So we found that x ≡ 7231021
If this is correct, then the following statements (i.e. the original equations) are true:
541x (mod 237) ≡ 637 (mod 237)
541x (mod 451) ≡ 694 (mod 451)
921x (mod 577) ≡ 567 (mod 577)

Let's see whether that's indeed the case if we use x ≡ 7231021.