Bootstrap
  C.R.T. .com
It doesn't have to be difficult if someone just explains it right.

Welcome to ChineseRemainderTheorem.com!

×

Modal Header

Some text in the Modal Body

Some other text...

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
33878033010
878332620101
332011301-1
2013171-12
13716-12-3
76112-35
6160-35-33
So our multiplicative inverse is 5 mod 33 ≡ 5
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
164001010
64016400101
So our multiplicative inverse is 0 mod 1 ≡ 0
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
89321742501-4
217258171-433
251718-433-37
1782133-37107
8180-37107-893
So our multiplicative inverse is 107 mod 893 ≡ 107
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 ≡ 670 × 878-1 (mod 33) ≡ 670 × 5 (mod 33) ≡ 17 (mod 33)
x ≡ 374 × 640-1 (mod 1) ≡ 374 × 0 (mod 1) ≡ 0 (mod 1)
x ≡ 192 × 217-1 (mod 893) ≡ 192 × 107 (mod 893) ≡ 5 (mod 893)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 33 × 1 × 893 = 29469
  2. We calculate the numbers M1 to M3
    M1=M/m1=29469/33=893,   M2=M/m2=29469/1=29469,   M3=M/m3=29469/893=33
  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
    33893033010
    89333272101
    33216101-16
    21201-1633
    So our multiplicative inverse is -16 mod 33 ≡ 17
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    12946901010
    294691294690101
    So our multiplicative inverse is 0 mod 1 ≡ 0
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    8933327201-27
    3321611-27433
    2120-27433-893
    So our multiplicative inverse is 433 mod 893 ≡ 433
    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
    =  (17 × 893 × 17 +
       0 × 29469 × 0 +
       5 × 33 × 433)   mod 29469
    = 5363 (mod 29469)


    So our answer is 5363 (mod 29469).


Verification

So we found that x ≡ 5363
If this is correct, then the following statements (i.e. the original equations) are true:
878x (mod 33) ≡ 670 (mod 33)
640x (mod 1) ≡ 374 (mod 1)
217x (mod 893) ≡ 192 (mod 893)

Let's see whether that's indeed the case if we use x ≡ 5363.