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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
1914750191010
475191293101
191932501-2
9351831-237
5312-237-39
321137-3976
2120-3976-191
So our multiplicative inverse is 76 mod 191 ≡ 76
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
614411701-1
44172101-13
171017-13-4
107133-47
7321-47-18
31307-1861
So our multiplicative inverse is -18 mod 61 ≡ 43
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1011450101010
145101144101
1014421301-2
4413351-27
13523-27-16
53127-1623
3211-1623-39
212023-39101
So our multiplicative inverse is -39 mod 101 ≡ 62
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 ≡ 334 × 475-1 (mod 191) ≡ 334 × 76 (mod 191) ≡ 172 (mod 191)
x ≡ 905 × 44-1 (mod 61) ≡ 905 × 43 (mod 61) ≡ 58 (mod 61)
x ≡ 701 × 145-1 (mod 101) ≡ 701 × 62 (mod 101) ≡ 32 (mod 101)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 191 × 61 × 101 = 1176751
  2. We calculate the numbers M1 to M3
    M1=M/m1=1176751/191=6161,   M2=M/m2=1176751/61=19291,   M3=M/m3=1176751/101=11651
  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
    19161610191010
    61611913249101
    1914934401-3
    4944151-34
    44584-34-35
    54114-3539
    4140-3539-191
    So our multiplicative inverse is 39 mod 191 ≡ 39
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    6119291061010
    192916131615101
    61154101-4
    1511501-461
    So our multiplicative inverse is -4 mod 61 ≡ 57
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    101116510101010
    1165110111536101
    1013622901-2
    3629171-23
    29741-23-14
    71703-14101
    So our multiplicative inverse is -14 mod 101 ≡ 87
    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
    =  (172 × 6161 × 39 +
       58 × 19291 × 57 +
       32 × 11651 × 87)   mod 1176751
    = 1037302 (mod 1176751)


    So our answer is 1037302 (mod 1176751).


Verification

So we found that x ≡ 1037302
If this is correct, then the following statements (i.e. the original equations) are true:
475x (mod 191) ≡ 334 (mod 191)
44x (mod 61) ≡ 905 (mod 61)
145x (mod 101) ≡ 701 (mod 101)

Let's see whether that's indeed the case if we use x ≡ 1037302.