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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
3916070391010
6073911216101
391216117501-1
2161751411-12
17541411-12-9
4111382-929
11813-929-38
832229-38105
3211-38105-143
2120105-143391
So our multiplicative inverse is -143 mod 391 ≡ 248
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
2294650229010
46522927101
229732501-32
75121-3233
5221-3233-98
212033-98229
So our multiplicative inverse is -98 mod 229 ≡ 131
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
4034300403010
430403127101
40327142501-14
2725121-1415
252121-1415-194
212015-194403
So our multiplicative inverse is -194 mod 403 ≡ 209
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 ≡ 718 × 607-1 (mod 391) ≡ 718 × 248 (mod 391) ≡ 159 (mod 391)
x ≡ 541 × 465-1 (mod 229) ≡ 541 × 131 (mod 229) ≡ 110 (mod 229)
x ≡ 220 × 430-1 (mod 403) ≡ 220 × 209 (mod 403) ≡ 38 (mod 403)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 391 × 229 × 403 = 36084217
  2. We calculate the numbers M1 to M3
    M1=M/m1=36084217/391=92287,   M2=M/m2=36084217/229=157573,   M3=M/m3=36084217/403=89539
  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
    391922870391010
    9228739123611101
    3911135601-35
    116151-3536
    6511-3536-71
    515036-71391
    So our multiplicative inverse is -71 mod 391 ≡ 320
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    2291575730229010
    15757322968821101
    22921101901-10
    2119121-1011
    19291-1011-109
    212011-109229
    So our multiplicative inverse is -109 mod 229 ≡ 120
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    403895390403010
    8953940322273101
    4037353801-5
    73381351-56
    383513-56-11
    3531126-11127
    3211-11127-138
    2120127-138403
    So our multiplicative inverse is -138 mod 403 ≡ 265
    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
    =  (159 × 92287 × 320 +
       110 × 157573 × 120 +
       38 × 89539 × 265)   mod 36084217
    = 27329886 (mod 36084217)


    So our answer is 27329886 (mod 36084217).


Verification

So we found that x ≡ 27329886
If this is correct, then the following statements (i.e. the original equations) are true:
607x (mod 391) ≡ 718 (mod 391)
465x (mod 229) ≡ 541 (mod 229)
430x (mod 403) ≡ 220 (mod 403)

Let's see whether that's indeed the case if we use x ≡ 27329886.