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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
863661120201-1
6612023551-14
20255337-14-13
55371184-1317
371821-1317-47
18118017-47863
So our multiplicative inverse is -47 mod 863 ≡ 816
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1997670199010
7671993170101
19917012901-1
170295251-16
292514-16-7
254616-748
4140-748-199
So our multiplicative inverse is 48 mod 199 ≡ 48
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
3293980329010
398329169101
3296945301-4
69531161-45
531635-45-19
165315-1962
5150-1962-329
So our multiplicative inverse is 62 mod 329 ≡ 62
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 ≡ 467 × 661-1 (mod 863) ≡ 467 × 816 (mod 863) ≡ 489 (mod 863)
x ≡ 145 × 767-1 (mod 199) ≡ 145 × 48 (mod 199) ≡ 194 (mod 199)
x ≡ 147 × 398-1 (mod 329) ≡ 147 × 62 (mod 329) ≡ 231 (mod 329)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 863 × 199 × 329 = 56501473
  2. We calculate the numbers M1 to M3
    M1=M/m1=56501473/863=65471,   M2=M/m2=56501473/199=283927,   M3=M/m3=56501473/329=171737
  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
    863654710863010
    6547186375746101
    863746111701-1
    7461176441-17
    11744229-17-15
    44291157-1522
    2915114-1522-37
    15141122-3759
    141140-3759-863
    So our multiplicative inverse is 59 mod 863 ≡ 59
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    1992839270199010
    2839271991426153101
    19915314601-1
    153463151-14
    461531-14-13
    1511504-13199
    So our multiplicative inverse is -13 mod 199 ≡ 186
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    3291717370329010
    171737329521328101
    3293281101-1
    328132801-1329
    So our multiplicative inverse is -1 mod 329 ≡ 328
    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
    =  (489 × 65471 × 59 +
       194 × 283927 × 186 +
       231 × 171737 × 328)   mod 56501473
    = 3139220 (mod 56501473)


    So our answer is 3139220 (mod 56501473).


Verification

So we found that x ≡ 3139220
If this is correct, then the following statements (i.e. the original equations) are true:
661x (mod 863) ≡ 467 (mod 863)
767x (mod 199) ≡ 145 (mod 199)
398x (mod 329) ≡ 147 (mod 329)

Let's see whether that's indeed the case if we use x ≡ 3139220.