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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
3537230353010
723353217101
35317201301-20
1713141-2021
13431-2021-83
414021-83353
So our multiplicative inverse is -83 mod 353 ≡ 270
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
49202901-2
209221-25
9241-25-22
21205-2249
So our multiplicative inverse is -22 mod 49 ≡ 27
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
7245144401-144
54111-144145
4140-144145-724
So our multiplicative inverse is 145 mod 724 ≡ 145
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 ≡ 566 × 723-1 (mod 353) ≡ 566 × 270 (mod 353) ≡ 324 (mod 353)
x ≡ 359 × 20-1 (mod 49) ≡ 359 × 27 (mod 49) ≡ 40 (mod 49)
x ≡ 941 × 5-1 (mod 724) ≡ 941 × 145 (mod 724) ≡ 333 (mod 724)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 353 × 49 × 724 = 12523028
  2. We calculate the numbers M1 to M3
    M1=M/m1=12523028/353=35476,   M2=M/m2=12523028/49=255572,   M3=M/m3=12523028/724=17297
  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
    353354760353010
    35476353100176101
    3531762101-2
    176117601-2353
    So our multiplicative inverse is -2 mod 353 ≡ 351
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    49255572049010
    25557249521537101
    493711201-1
    3712311-14
    121120-14-49
    So our multiplicative inverse is 4 mod 49 ≡ 4
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    724172970724010
    1729772423645101
    72464517901-1
    645798131-19
    791361-19-55
    1311309-55724
    So our multiplicative inverse is -55 mod 724 ≡ 669
    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
    =  (324 × 35476 × 351 +
       40 × 255572 × 4 +
       333 × 17297 × 669)   mod 12523028
    = 1661189 (mod 12523028)


    So our answer is 1661189 (mod 12523028).


Verification

So we found that x ≡ 1661189
If this is correct, then the following statements (i.e. the original equations) are true:
723x (mod 353) ≡ 566 (mod 353)
20x (mod 49) ≡ 359 (mod 49)
5x (mod 724) ≡ 941 (mod 724)

Let's see whether that's indeed the case if we use x ≡ 1661189.