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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
2692611801-1
26183251-133
8513-133-34
531233-3467
3211-3467-101
212067-101269
So our multiplicative inverse is -101 mod 269 ≡ 168
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1905390190010
5391902159101
19015913101-1
15931541-16
31473-16-43
43116-4349
3130-4349-190
So our multiplicative inverse is 49 mod 190 ≡ 49
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
61922061010
92261157101
6178501-8
75121-89
5221-89-26
21209-2661
So our multiplicative inverse is -26 mod 61 ≡ 35
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 ≡ 290 × 261-1 (mod 269) ≡ 290 × 168 (mod 269) ≡ 31 (mod 269)
x ≡ 610 × 539-1 (mod 190) ≡ 610 × 49 (mod 190) ≡ 60 (mod 190)
x ≡ 49 × 922-1 (mod 61) ≡ 49 × 35 (mod 61) ≡ 7 (mod 61)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 269 × 190 × 61 = 3117710
  2. We calculate the numbers M1 to M3
    M1=M/m1=3117710/269=11590,   M2=M/m2=3117710/190=16409,   M3=M/m3=3117710/61=51110
  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
    269115900269010
    115902694323101
    26923111601-11
    2316171-1112
    16722-1112-35
    723112-35117
    2120-35117-269
    So our multiplicative inverse is 117 mod 269 ≡ 117
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    190164090190010
    164091908669101
    1906925201-2
    69521171-23
    521731-23-11
    1711703-11190
    So our multiplicative inverse is -11 mod 190 ≡ 179
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    6151110061010
    511106183753101
    61531801-1
    538651-17
    8513-17-8
    53127-815
    3211-815-23
    212015-2361
    So our multiplicative inverse is -23 mod 61 ≡ 38
    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
    =  (31 × 11590 × 117 +
       60 × 16409 × 179 +
       7 × 51110 × 38)   mod 3117710
    = 1154310 (mod 3117710)


    So our answer is 1154310 (mod 3117710).


Verification

So we found that x ≡ 1154310
If this is correct, then the following statements (i.e. the original equations) are true:
261x (mod 269) ≡ 290 (mod 269)
539x (mod 190) ≡ 610 (mod 190)
922x (mod 61) ≡ 49 (mod 61)

Let's see whether that's indeed the case if we use x ≡ 1154310.