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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
11776011010
77611706101
1161501-1
65111-12
5150-12-11
So our multiplicative inverse is 2 mod 11 ≡ 2
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
6197185101-8
71511201-89
5120211-89-26
2011199-2635
11912-2635-61
924135-61279
2120-61279-619
So our multiplicative inverse is 279 mod 619 ≡ 279
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
563197216901-2
1971691281-23
1692861-23-20
2812803-20563
So our multiplicative inverse is -20 mod 563 ≡ 543
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 ≡ 401 × 776-1 (mod 11) ≡ 401 × 2 (mod 11) ≡ 10 (mod 11)
x ≡ 160 × 71-1 (mod 619) ≡ 160 × 279 (mod 619) ≡ 72 (mod 619)
x ≡ 20 × 197-1 (mod 563) ≡ 20 × 543 (mod 563) ≡ 163 (mod 563)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 11 × 619 × 563 = 3833467
  2. We calculate the numbers M1 to M3
    M1=M/m1=3833467/11=348497,   M2=M/m2=3833467/619=6193,   M3=M/m3=3833467/563=6809
  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
    11348497011010
    34849711316816101
    1161501-1
    65111-12
    5150-12-11
    So our multiplicative inverse is 2 mod 11 ≡ 2
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    61961930619010
    6193619103101
    6193206101-206
    31301-206619
    So our multiplicative inverse is -206 mod 619 ≡ 413
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    56368090563010
    68095631253101
    56353103301-10
    53331201-1011
    3320113-1011-21
    20131711-2132
    13716-2132-53
    761132-5385
    6160-5385-563
    So our multiplicative inverse is 85 mod 563 ≡ 85
    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
    =  (10 × 348497 × 2 +
       72 × 6193 × 413 +
       163 × 6809 × 85)   mod 3833467
    = 1787125 (mod 3833467)


    So our answer is 1787125 (mod 3833467).


Verification

So we found that x ≡ 1787125
If this is correct, then the following statements (i.e. the original equations) are true:
776x (mod 11) ≡ 401 (mod 11)
71x (mod 619) ≡ 160 (mod 619)
197x (mod 563) ≡ 20 (mod 563)

Let's see whether that's indeed the case if we use x ≡ 1787125.