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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
71664071010
66471925101
712522101-2
2521141-23
21451-23-17
41403-1771
So our multiplicative inverse is -17 mod 71 ≡ 54
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
85504085010
50485579101
85791601-1
7961311-114
6160-114-85
So our multiplicative inverse is 14 mod 85 ≡ 14
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
4434371601-1
43767251-173
6511-173-74
515073-74443
So our multiplicative inverse is -74 mod 443 ≡ 369
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 ≡ 242 × 664-1 (mod 71) ≡ 242 × 54 (mod 71) ≡ 4 (mod 71)
x ≡ 690 × 504-1 (mod 85) ≡ 690 × 14 (mod 85) ≡ 55 (mod 85)
x ≡ 834 × 437-1 (mod 443) ≡ 834 × 369 (mod 443) ≡ 304 (mod 443)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 71 × 85 × 443 = 2673505
  2. We calculate the numbers M1 to M3
    M1=M/m1=2673505/71=37655,   M2=M/m2=2673505/85=31453,   M3=M/m3=2673505/443=6035
  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
    7137655071010
    376557153025101
    712522101-2
    2521141-23
    21451-23-17
    41403-1771
    So our multiplicative inverse is -17 mod 71 ≡ 54
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    8531453085010
    31453853703101
    85328101-28
    31301-2885
    So our multiplicative inverse is -28 mod 85 ≡ 57
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    44360350443010
    603544313276101
    443276116701-1
    27616711091-12
    167109158-12-3
    109581512-35
    585117-35-8
    517725-861
    7231-861-191
    212061-191443
    So our multiplicative inverse is -191 mod 443 ≡ 252
    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
    =  (4 × 37655 × 54 +
       55 × 31453 × 57 +
       304 × 6035 × 252)   mod 2673505
    = 2284855 (mod 2673505)


    So our answer is 2284855 (mod 2673505).


Verification

So we found that x ≡ 2284855
If this is correct, then the following statements (i.e. the original equations) are true:
664x (mod 71) ≡ 242 (mod 71)
504x (mod 85) ≡ 690 (mod 85)
437x (mod 443) ≡ 834 (mod 443)

Let's see whether that's indeed the case if we use x ≡ 2284855.