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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
1132600113010
260113234101
1133431101-3
3411311-310
111110-310-113
So our multiplicative inverse is 10 mod 113 ≡ 10
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
8678601701-1
860712261-1123
7611-1123-124
6160123-124867
So our multiplicative inverse is -124 mod 867 ≡ 743
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
4229650422010
9654222121101
42212135901-3
12159231-37
593192-37-136
32117-136143
2120-136143-422
So our multiplicative inverse is 143 mod 422 ≡ 143
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 ≡ 392 × 260-1 (mod 113) ≡ 392 × 10 (mod 113) ≡ 78 (mod 113)
x ≡ 675 × 860-1 (mod 867) ≡ 675 × 743 (mod 867) ≡ 399 (mod 867)
x ≡ 944 × 965-1 (mod 422) ≡ 944 × 143 (mod 422) ≡ 374 (mod 422)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 113 × 867 × 422 = 41343762
  2. We calculate the numbers M1 to M3
    M1=M/m1=41343762/113=365874,   M2=M/m2=41343762/867=47686,   M3=M/m3=41343762/422=97971
  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
    1133658740113010
    365874113323793101
    1139312001-1
    93204131-15
    201317-15-6
    137165-611
    7611-611-17
    616011-17113
    So our multiplicative inverse is -17 mod 113 ≡ 96
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    867476860867010
    47686867551101
    8671867001-867
    So our multiplicative inverse is 1 mod 867 ≡ 1
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    422979710422010
    9797142223267101
    4226762001-6
    6720371-619
    20726-619-44
    761119-4463
    6160-4463-422
    So our multiplicative inverse is 63 mod 422 ≡ 63
    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
    =  (78 × 365874 × 96 +
       399 × 47686 × 1 +
       374 × 97971 × 63)   mod 41343762
    = 23144964 (mod 41343762)


    So our answer is 23144964 (mod 41343762).


Verification

So we found that x ≡ 23144964
If this is correct, then the following statements (i.e. the original equations) are true:
260x (mod 113) ≡ 392 (mod 113)
860x (mod 867) ≡ 675 (mod 867)
965x (mod 422) ≡ 944 (mod 422)

Let's see whether that's indeed the case if we use x ≡ 23144964.