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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
9915198101-198
51501-198991
So our multiplicative inverse is -198 mod 991 ≡ 793
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
3273770327010
377327150101
3275062701-6
50271231-67
272314-67-13
234537-1372
4311-1372-85
313072-85327
So our multiplicative inverse is -85 mod 327 ≡ 242
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
55623728201-2
237822731-25
827319-25-7
739815-761
9190-761-556
So our multiplicative inverse is 61 mod 556 ≡ 61
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 ≡ 203 × 5-1 (mod 991) ≡ 203 × 793 (mod 991) ≡ 437 (mod 991)
x ≡ 746 × 377-1 (mod 327) ≡ 746 × 242 (mod 327) ≡ 28 (mod 327)
x ≡ 152 × 237-1 (mod 556) ≡ 152 × 61 (mod 556) ≡ 376 (mod 556)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 991 × 327 × 556 = 180175692
  2. We calculate the numbers M1 to M3
    M1=M/m1=180175692/991=181812,   M2=M/m2=180175692/327=550996,   M3=M/m3=180175692/556=324057
  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
    9911818120991010
    181812991183459101
    99145927301-2
    459736211-213
    7321310-213-41
    21102113-4195
    101100-4195-991
    So our multiplicative inverse is 95 mod 991 ≡ 95
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    3275509960327010
    55099632716851101
    3271327001-327
    So our multiplicative inverse is 1 mod 327 ≡ 1
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    5563240570556010
    324057556582465101
    55646519101-1
    465915101-16
    911091-16-55
    1011006-55556
    So our multiplicative inverse is -55 mod 556 ≡ 501
    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
    =  (437 × 181812 × 95 +
       28 × 550996 × 1 +
       376 × 324057 × 501)   mod 180175692
    = 141151540 (mod 180175692)


    So our answer is 141151540 (mod 180175692).


Verification

So we found that x ≡ 141151540
If this is correct, then the following statements (i.e. the original equations) are true:
5x (mod 991) ≡ 203 (mod 991)
377x (mod 327) ≡ 746 (mod 327)
237x (mod 556) ≡ 152 (mod 556)

Let's see whether that's indeed the case if we use x ≡ 141151540.