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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
427269115801-1
26915811111-12
158111147-12-3
111472172-38
4717213-38-19
1713148-1927
13431-1927-100
414027-100427
So our multiplicative inverse is -100 mod 427 ≡ 327
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
7668610766010
861766195101
766958601-8
9561551-8121
6511-8121-129
5150121-129766
So our multiplicative inverse is -129 mod 766 ≡ 637
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1438390143010
8391435124101
14312411901-1
124196101-17
191019-17-8
109117-815
9190-815-143
So our multiplicative inverse is 15 mod 143 ≡ 15
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 ≡ 673 × 269-1 (mod 427) ≡ 673 × 327 (mod 427) ≡ 166 (mod 427)
x ≡ 198 × 861-1 (mod 766) ≡ 198 × 637 (mod 766) ≡ 502 (mod 766)
x ≡ 373 × 839-1 (mod 143) ≡ 373 × 15 (mod 143) ≡ 18 (mod 143)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 427 × 766 × 143 = 46772726
  2. We calculate the numbers M1 to M3
    M1=M/m1=46772726/427=109538,   M2=M/m2=46772726/766=61061,   M3=M/m3=46772726/143=327082
  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
    4271095380427010
    109538427256226101
    427226120101-1
    2262011251-12
    2012581-12-17
    2512502-17427
    So our multiplicative inverse is -17 mod 427 ≡ 410
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    766610610766010
    6106176679547101
    766547121901-1
    54721921091-13
    21910921-13-7
    109110903-7766
    So our multiplicative inverse is -7 mod 766 ≡ 759
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    1433270820143010
    327082143228741101
    1434132001-3
    4120211-37
    201200-37-143
    So our multiplicative inverse is 7 mod 143 ≡ 7
    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
    =  (166 × 109538 × 410 +
       502 × 61061 × 759 +
       18 × 327082 × 7)   mod 46772726
    = 32027728 (mod 46772726)


    So our answer is 32027728 (mod 46772726).


Verification

So we found that x ≡ 32027728
If this is correct, then the following statements (i.e. the original equations) are true:
269x (mod 427) ≡ 673 (mod 427)
861x (mod 766) ≡ 198 (mod 766)
839x (mod 143) ≡ 373 (mod 143)

Let's see whether that's indeed the case if we use x ≡ 32027728.