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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
295502010
95524771101
212001-2
So our multiplicative inverse is 1 mod 2 ≡ 1
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
5115150511010
51551114101
5114127301-127
43111-127128
3130-127128-511
So our multiplicative inverse is 128 mod 511 ≡ 128
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
64128327501-2
283753581-27
7558117-27-9
5817377-934
17723-934-77
732134-77188
3130-77188-641
So our multiplicative inverse is 188 mod 641 ≡ 188
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 ≡ 105 × 955-1 (mod 2) ≡ 105 × 1 (mod 2) ≡ 1 (mod 2)
x ≡ 417 × 515-1 (mod 511) ≡ 417 × 128 (mod 511) ≡ 232 (mod 511)
x ≡ 629 × 283-1 (mod 641) ≡ 629 × 188 (mod 641) ≡ 308 (mod 641)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 2 × 511 × 641 = 655102
  2. We calculate the numbers M1 to M3
    M1=M/m1=655102/2=327551,   M2=M/m2=655102/511=1282,   M3=M/m3=655102/641=1022
  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
    232755102010
    32755121637751101
    212001-2
    So our multiplicative inverse is 1 mod 2 ≡ 1
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    51112820511010
    12825112260101
    511260125101-1
    260251191-12
    2519278-12-55
    98112-5557
    8180-5557-511
    So our multiplicative inverse is 57 mod 511 ≡ 57
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    64110220641010
    10226411381101
    641381126001-1
    38126011211-12
    260121218-12-5
    121186132-532
    181315-532-37
    1352332-37106
    5312-37106-143
    3211106-143249
    2120-143249-641
    So our multiplicative inverse is 249 mod 641 ≡ 249
    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
    =  (1 × 327551 × 1 +
       232 × 1282 × 57 +
       308 × 1022 × 249)   mod 655102
    = 15051 (mod 655102)


    So our answer is 15051 (mod 655102).


Verification

So we found that x ≡ 15051
If this is correct, then the following statements (i.e. the original equations) are true:
955x (mod 2) ≡ 105 (mod 2)
515x (mod 511) ≡ 417 (mod 511)
283x (mod 641) ≡ 629 (mod 641)

Let's see whether that's indeed the case if we use x ≡ 15051.