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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
1538600153010
860153595101
1539515801-1
95581371-12
5837121-12-3
37211162-35
211615-35-8
165315-829
5150-829-153
So our multiplicative inverse is 29 mod 153 ≡ 29
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
98353182901-18
53291241-1819
292415-1819-37
2454419-37167
5411-37167-204
4140167-204983
So our multiplicative inverse is -204 mod 983 ≡ 779
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
4437030443010
7034431260101
443260118301-1
2601831771-12
18377229-12-5
77292192-512
2919110-512-17
19101912-1729
10911-1729-46
919029-46443
So our multiplicative inverse is -46 mod 443 ≡ 397
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 ≡ 455 × 860-1 (mod 153) ≡ 455 × 29 (mod 153) ≡ 37 (mod 153)
x ≡ 896 × 53-1 (mod 983) ≡ 896 × 779 (mod 983) ≡ 54 (mod 983)
x ≡ 193 × 703-1 (mod 443) ≡ 193 × 397 (mod 443) ≡ 425 (mod 443)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 153 × 983 × 443 = 66626757
  2. We calculate the numbers M1 to M3
    M1=M/m1=66626757/153=435469,   M2=M/m2=66626757/983=67779,   M3=M/m3=66626757/443=150399
  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
    1534354690153010
    435469153284631101
    1533142901-4
    3129121-45
    292141-45-74
    21205-74153
    So our multiplicative inverse is -74 mod 153 ≡ 79
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    983677790983010
    6777998368935101
    98393514801-1
    9354819231-120
    482322-120-41
    23211120-41471
    2120-41471-983
    So our multiplicative inverse is 471 mod 983 ≡ 471
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    4431503990443010
    150399443339222101
    443222122101-1
    222221111-12
    22112210-12-443
    So our multiplicative inverse is 2 mod 443 ≡ 2
    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
    =  (37 × 435469 × 79 +
       54 × 67779 × 471 +
       425 × 150399 × 2)   mod 66626757
    = 59775301 (mod 66626757)


    So our answer is 59775301 (mod 66626757).


Verification

So we found that x ≡ 59775301
If this is correct, then the following statements (i.e. the original equations) are true:
860x (mod 153) ≡ 455 (mod 153)
53x (mod 983) ≡ 896 (mod 983)
703x (mod 443) ≡ 193 (mod 443)

Let's see whether that's indeed the case if we use x ≡ 59775301.