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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
919331225701-2
3312571741-23
25774335-23-11
7435243-1125
35483-1125-211
431125-211236
3130-211236-919
So our multiplicative inverse is 236 mod 919 ≡ 236
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
41884041010
884412123101
412311801-1
2318151-12
18533-12-7
53122-79
3211-79-16
21209-1641
So our multiplicative inverse is -16 mod 41 ≡ 25
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
9896164501-164
65111-164165
5150-164165-989
So our multiplicative inverse is 165 mod 989 ≡ 165
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 ≡ 949 × 331-1 (mod 919) ≡ 949 × 236 (mod 919) ≡ 647 (mod 919)
x ≡ 564 × 884-1 (mod 41) ≡ 564 × 25 (mod 41) ≡ 37 (mod 41)
x ≡ 362 × 6-1 (mod 989) ≡ 362 × 165 (mod 989) ≡ 390 (mod 989)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 919 × 41 × 989 = 37264531
  2. We calculate the numbers M1 to M3
    M1=M/m1=37264531/919=40549,   M2=M/m2=37264531/41=908891,   M3=M/m3=37264531/989=37679
  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
    919405490919010
    4054991944113101
    91911381501-8
    11315781-857
    15817-857-65
    871157-65122
    7170-65122-919
    So our multiplicative inverse is 122 mod 919 ≡ 122
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    41908891041010
    90889141221683101
    41313201-13
    32111-1314
    2120-1314-41
    So our multiplicative inverse is 14 mod 41 ≡ 14
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    989376790989010
    376799893897101
    98997101901-10
    9719521-1051
    19291-1051-469
    212051-469989
    So our multiplicative inverse is -469 mod 989 ≡ 520
    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
    =  (647 × 40549 × 122 +
       37 × 908891 × 14 +
       390 × 37679 × 520)   mod 37264531
    = 21648611 (mod 37264531)


    So our answer is 21648611 (mod 37264531).


Verification

So we found that x ≡ 21648611
If this is correct, then the following statements (i.e. the original equations) are true:
331x (mod 919) ≡ 949 (mod 919)
884x (mod 41) ≡ 564 (mod 41)
6x (mod 989) ≡ 362 (mod 989)

Let's see whether that's indeed the case if we use x ≡ 21648611.