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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
52487052010
48752919101
521921401-2
1914151-23
14524-23-8
54113-811
4140-811-52
So our multiplicative inverse is 11 mod 52 ≡ 11
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
785466131901-1
46631911471-12
319147225-12-5
147255222-527
252213-527-32
2237127-32251
3130-32251-785
So our multiplicative inverse is 251 mod 785 ≡ 251
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
53379053010
3795378101
5386501-6
85131-67
5312-67-13
32117-1320
2120-1320-53
So our multiplicative inverse is 20 mod 53 ≡ 20
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 ≡ 99 × 487-1 (mod 52) ≡ 99 × 11 (mod 52) ≡ 49 (mod 52)
x ≡ 307 × 466-1 (mod 785) ≡ 307 × 251 (mod 785) ≡ 127 (mod 785)
x ≡ 425 × 379-1 (mod 53) ≡ 425 × 20 (mod 53) ≡ 20 (mod 53)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 52 × 785 × 53 = 2163460
  2. We calculate the numbers M1 to M3
    M1=M/m1=2163460/52=41605,   M2=M/m2=2163460/785=2756,   M3=M/m3=2163460/53=40820
  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
    5241605052010
    41605528005101
    52510201-10
    52211-1021
    2120-1021-52
    So our multiplicative inverse is 21 mod 52 ≡ 21
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    78527560785010
    27567853401101
    785401138401-1
    4013841171-12
    384172210-12-45
    1710172-4547
    10713-4547-92
    732147-92231
    3130-92231-785
    So our multiplicative inverse is 231 mod 785 ≡ 231
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    5340820053010
    408205377010101
    53105301-5
    103311-516
    3130-516-53
    So our multiplicative inverse is 16 mod 53 ≡ 16
    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
    =  (49 × 41605 × 21 +
       127 × 2756 × 231 +
       20 × 40820 × 16)   mod 2163460
    = 428737 (mod 2163460)


    So our answer is 428737 (mod 2163460).


Verification

So we found that x ≡ 428737
If this is correct, then the following statements (i.e. the original equations) are true:
487x (mod 52) ≡ 99 (mod 52)
466x (mod 785) ≡ 307 (mod 785)
379x (mod 53) ≡ 425 (mod 53)

Let's see whether that's indeed the case if we use x ≡ 428737.