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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
829705112401-1
7051245851-16
12485139-16-7
8539276-720
39754-720-107
741320-107127
4311-107127-234
3130127-234829
So our multiplicative inverse is -234 mod 829 ≡ 595
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
3897500389010
7503891361101
38936112801-1
3612812251-113
282513-113-14
2538113-14125
3130-14125-389
So our multiplicative inverse is 125 mod 389 ≡ 125
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
3825371101-7
5311491-729
11912-729-36
924129-36173
2120-36173-382
So our multiplicative inverse is 173 mod 382 ≡ 173
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 ≡ 234 × 705-1 (mod 829) ≡ 234 × 595 (mod 829) ≡ 787 (mod 829)
x ≡ 822 × 750-1 (mod 389) ≡ 822 × 125 (mod 389) ≡ 54 (mod 389)
x ≡ 50 × 53-1 (mod 382) ≡ 50 × 173 (mod 382) ≡ 246 (mod 382)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 829 × 389 × 382 = 123187742
  2. We calculate the numbers M1 to M3
    M1=M/m1=123187742/829=148598,   M2=M/m2=123187742/389=316678,   M3=M/m3=123187742/382=322481
  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
    8291485980829010
    148598829179207101
    8292074101-4
    207120701-4829
    So our multiplicative inverse is -4 mod 829 ≡ 825
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    3893166780389010
    31667838981432101
    3893212501-12
    325621-1273
    5221-1273-158
    212073-158389
    So our multiplicative inverse is -158 mod 389 ≡ 231
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    3823224810382010
    32248138284473101
    3827351701-5
    7317451-521
    17532-521-68
    522121-68157
    2120-68157-382
    So our multiplicative inverse is 157 mod 382 ≡ 157
    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
    =  (787 × 148598 × 825 +
       54 × 316678 × 231 +
       246 × 322481 × 157)   mod 123187742
    = 46097332 (mod 123187742)


    So our answer is 46097332 (mod 123187742).


Verification

So we found that x ≡ 46097332
If this is correct, then the following statements (i.e. the original equations) are true:
705x (mod 829) ≡ 234 (mod 829)
750x (mod 389) ≡ 822 (mod 389)
53x (mod 382) ≡ 50 (mod 382)

Let's see whether that's indeed the case if we use x ≡ 46097332.