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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
42734817901-1
348794321-15
7932215-15-11
3215225-1127
15271-1127-200
212027-200427
So our multiplicative inverse is -200 mod 427 ≡ 227
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
8669290866010
929866163101
86663134701-13
63471161-1314
4716215-1314-41
16151114-4155
151150-4155-866
So our multiplicative inverse is 55 mod 866 ≡ 55
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
783649113401-1
64913441131-15
134113121-15-6
11321585-635
21825-635-76
851335-76111
5312-76111-187
3211111-187298
2120-187298-783
So our multiplicative inverse is 298 mod 783 ≡ 298
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 ≡ 498 × 348-1 (mod 427) ≡ 498 × 227 (mod 427) ≡ 318 (mod 427)
x ≡ 946 × 929-1 (mod 866) ≡ 946 × 55 (mod 866) ≡ 70 (mod 866)
x ≡ 382 × 649-1 (mod 783) ≡ 382 × 298 (mod 783) ≡ 301 (mod 783)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 427 × 866 × 783 = 289539306
  2. We calculate the numbers M1 to M3
    M1=M/m1=289539306/427=678078,   M2=M/m2=289539306/866=334341,   M3=M/m3=289539306/783=369782
  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
    4276780780427010
    67807842715882101
    4272213101-213
    21201-213427
    So our multiplicative inverse is -213 mod 427 ≡ 214
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    8663343410866010
    33434186638665101
    86665132101-13
    6521321-1340
    212101-1340-413
    212040-413866
    So our multiplicative inverse is -413 mod 866 ≡ 453
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    7833697820783010
    369782783472206101
    783206316501-3
    2061651411-34
    1654141-34-19
    4114104-19783
    So our multiplicative inverse is -19 mod 783 ≡ 764
    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
    =  (318 × 678078 × 214 +
       70 × 334341 × 453 +
       301 × 369782 × 764)   mod 289539306
    = 198344380 (mod 289539306)


    So our answer is 198344380 (mod 289539306).


Verification

So we found that x ≡ 198344380
If this is correct, then the following statements (i.e. the original equations) are true:
348x (mod 427) ≡ 498 (mod 427)
929x (mod 866) ≡ 946 (mod 866)
649x (mod 783) ≡ 382 (mod 783)

Let's see whether that's indeed the case if we use x ≡ 198344380.