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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
40619322001-2
193209131-219
201317-219-21
1371619-2140
7611-2140-61
616040-61406
So our multiplicative inverse is -61 mod 406 ≡ 345
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
313202111101-1
2021111911-12
11191120-12-3
91204112-314
201119-314-17
1191214-1731
9241-1731-141
212031-141313
So our multiplicative inverse is -141 mod 313 ≡ 172
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
47361047010
36147732101
473211501-1
3215221-13
15271-13-22
21203-2247
So our multiplicative inverse is -22 mod 47 ≡ 25
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 ≡ 583 × 193-1 (mod 406) ≡ 583 × 345 (mod 406) ≡ 165 (mod 406)
x ≡ 351 × 202-1 (mod 313) ≡ 351 × 172 (mod 313) ≡ 276 (mod 313)
x ≡ 832 × 361-1 (mod 47) ≡ 832 × 25 (mod 47) ≡ 26 (mod 47)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 406 × 313 × 47 = 5972666
  2. We calculate the numbers M1 to M3
    M1=M/m1=5972666/406=14711,   M2=M/m2=5972666/313=19082,   M3=M/m3=5972666/47=127078
  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
    406147110406010
    147114063695101
    4069542601-4
    95263171-413
    261719-413-17
    1791813-1730
    9811-1730-47
    818030-47406
    So our multiplicative inverse is -47 mod 406 ≡ 359
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    313190820313010
    1908231360302101
    31330211101-1
    302112751-128
    11521-128-57
    515028-57313
    So our multiplicative inverse is -57 mod 313 ≡ 256
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    47127078047010
    12707847270337101
    473711001-1
    3710371-14
    10713-14-5
    73214-514
    3130-514-47
    So our multiplicative inverse is 14 mod 47 ≡ 14
    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
    =  (165 × 14711 × 359 +
       276 × 19082 × 256 +
       26 × 127078 × 14)   mod 5972666
    = 2279855 (mod 5972666)


    So our answer is 2279855 (mod 5972666).


Verification

So we found that x ≡ 2279855
If this is correct, then the following statements (i.e. the original equations) are true:
193x (mod 406) ≡ 583 (mod 406)
202x (mod 313) ≡ 351 (mod 313)
361x (mod 47) ≡ 832 (mod 47)

Let's see whether that's indeed the case if we use x ≡ 2279855.