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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
763533123001-1
5332302731-13
23073311-13-10
7311673-1063
11714-1063-73
741363-73136
4311-73136-209
3130136-209763
So our multiplicative inverse is -209 mod 763 ≡ 554
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
4635050463010
505463142101
4634211101-11
4214201-11463
So our multiplicative inverse is -11 mod 463 ≡ 452
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
811664114701-1
6641474761-15
14776171-15-6
7671155-611
715141-611-160
515011-160811
So our multiplicative inverse is -160 mod 811 ≡ 651
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 ≡ 296 × 533-1 (mod 763) ≡ 296 × 554 (mod 763) ≡ 702 (mod 763)
x ≡ 397 × 505-1 (mod 463) ≡ 397 × 452 (mod 463) ≡ 263 (mod 463)
x ≡ 212 × 664-1 (mod 811) ≡ 212 × 651 (mod 811) ≡ 142 (mod 811)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 763 × 463 × 811 = 286501159
  2. We calculate the numbers M1 to M3
    M1=M/m1=286501159/763=375493,   M2=M/m2=286501159/463=618793,   M3=M/m3=286501159/811=353269
  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
    7633754930763010
    37549376349297101
    7639778401-7
    97841131-78
    841366-78-55
    136218-55118
    6160-55118-763
    So our multiplicative inverse is 118 mod 763 ≡ 118
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    4636187930463010
    6187934631336225101
    46322521301-2
    225131741-235
    13431-235-107
    414035-107463
    So our multiplicative inverse is -107 mod 463 ≡ 356
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    8113532690811010
    353269811435484101
    811484132701-1
    48432711571-12
    327157213-12-5
    157131212-562
    131130-562-811
    So our multiplicative inverse is 62 mod 811 ≡ 62
    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
    =  (702 × 375493 × 118 +
       263 × 618793 × 356 +
       142 × 353269 × 62)   mod 286501159
    = 183997389 (mod 286501159)


    So our answer is 183997389 (mod 286501159).


Verification

So we found that x ≡ 183997389
If this is correct, then the following statements (i.e. the original equations) are true:
533x (mod 763) ≡ 296 (mod 763)
505x (mod 463) ≡ 397 (mod 463)
664x (mod 811) ≡ 212 (mod 811)

Let's see whether that's indeed the case if we use x ≡ 183997389.