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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
1638740163010
874163559101
1635924501-2
59451141-23
451433-23-11
143423-1147
3211-1147-58
212047-58163
So our multiplicative inverse is -58 mod 163 ≡ 105
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
43141112001-1
4112020111-121
201119-121-22
1191221-2243
9241-2243-194
212043-194431
So our multiplicative inverse is -194 mod 431 ≡ 237
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
4038760403010
876403270101
4037055301-5
70531171-56
531732-56-23
172816-23190
2120-23190-403
So our multiplicative inverse is 190 mod 403 ≡ 190
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 ≡ 671 × 874-1 (mod 163) ≡ 671 × 105 (mod 163) ≡ 39 (mod 163)
x ≡ 219 × 411-1 (mod 431) ≡ 219 × 237 (mod 431) ≡ 183 (mod 431)
x ≡ 823 × 876-1 (mod 403) ≡ 823 × 190 (mod 403) ≡ 6 (mod 403)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 163 × 431 × 403 = 28311959
  2. We calculate the numbers M1 to M3
    M1=M/m1=28311959/163=173693,   M2=M/m2=28311959/431=65689,   M3=M/m3=28311959/403=70253
  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
    1631736930163010
    173693163106598101
    1639816501-1
    98651331-12
    6533132-12-3
    3332112-35
    321320-35-163
    So our multiplicative inverse is 5 mod 163 ≡ 5
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    431656890431010
    65689431152177101
    43117727701-2
    177772231-25
    772338-25-17
    238275-1739
    8711-1739-56
    717039-56431
    So our multiplicative inverse is -56 mod 431 ≡ 375
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    403702530403010
    70253403174131101
    40313131001-3
    131101311-340
    101100-340-403
    So our multiplicative inverse is 40 mod 403 ≡ 40
    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
    =  (39 × 173693 × 5 +
       183 × 65689 × 375 +
       6 × 70253 × 40)   mod 28311959
    = 413081 (mod 28311959)


    So our answer is 413081 (mod 28311959).


Verification

So we found that x ≡ 413081
If this is correct, then the following statements (i.e. the original equations) are true:
874x (mod 163) ≡ 671 (mod 163)
411x (mod 431) ≡ 219 (mod 431)
876x (mod 403) ≡ 823 (mod 403)

Let's see whether that's indeed the case if we use x ≡ 413081.