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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
5718340571010
8345711263101
57126324501-2
263455381-211
453817-211-13
3875311-1376
7321-1376-165
313076-165571
So our multiplicative inverse is -165 mod 571 ≡ 406
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
67341067010
3416756101
67611101-11
61601-1167
So our multiplicative inverse is -11 mod 67 ≡ 56
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
6617980661010
7986611137101
661137411301-4
1371131241-45
11324417-45-24
2417175-2429
17723-2429-82
732129-82193
3130-82193-661
So our multiplicative inverse is 193 mod 661 ≡ 193
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 ≡ 206 × 834-1 (mod 571) ≡ 206 × 406 (mod 571) ≡ 270 (mod 571)
x ≡ 734 × 341-1 (mod 67) ≡ 734 × 56 (mod 67) ≡ 33 (mod 67)
x ≡ 893 × 798-1 (mod 661) ≡ 893 × 193 (mod 661) ≡ 489 (mod 661)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 571 × 67 × 661 = 25287877
  2. We calculate the numbers M1 to M3
    M1=M/m1=25287877/571=44287,   M2=M/m2=25287877/67=377431,   M3=M/m3=25287877/661=38257
  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
    571442870571010
    4428757177320101
    571320125101-1
    3202511691-12
    25169344-12-7
    69441252-79
    4425119-79-16
    2519169-1625
    19631-1625-91
    616025-91571
    So our multiplicative inverse is -91 mod 571 ≡ 480
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    67377431067010
    37743167563320101
    67203701-3
    207261-37
    7611-37-10
    61607-1067
    So our multiplicative inverse is -10 mod 67 ≡ 57
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    661382570661010
    3825766157580101
    66158018101-1
    580817131-18
    811363-18-49
    133418-49204
    3130-49204-661
    So our multiplicative inverse is 204 mod 661 ≡ 204
    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
    =  (270 × 44287 × 480 +
       33 × 377431 × 57 +
       489 × 38257 × 204)   mod 25287877
    = 24318018 (mod 25287877)


    So our answer is 24318018 (mod 25287877).


Verification

So we found that x ≡ 24318018
If this is correct, then the following statements (i.e. the original equations) are true:
834x (mod 571) ≡ 206 (mod 571)
341x (mod 67) ≡ 734 (mod 67)
798x (mod 661) ≡ 893 (mod 661)

Let's see whether that's indeed the case if we use x ≡ 24318018.