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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
8188131501-1
813516231-1163
5312-1163-164
3211163-164327
2120-164327-818
So our multiplicative inverse is 327 mod 818 ≡ 327
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
525206211301-2
2061131931-23
11393120-23-5
93204133-523
201317-523-28
1371623-2851
7611-2851-79
616051-79525
So our multiplicative inverse is -79 mod 525 ≡ 446
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1312630131010
26313121101
1311131001-131
So our multiplicative inverse is 1 mod 131 ≡ 1
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 ≡ 23 × 813-1 (mod 818) ≡ 23 × 327 (mod 818) ≡ 159 (mod 818)
x ≡ 875 × 206-1 (mod 525) ≡ 875 × 446 (mod 525) ≡ 175 (mod 525)
x ≡ 430 × 263-1 (mod 131) ≡ 430 × 1 (mod 131) ≡ 37 (mod 131)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 818 × 525 × 131 = 56257950
  2. We calculate the numbers M1 to M3
    M1=M/m1=56257950/818=68775,   M2=M/m2=56257950/525=107158,   M3=M/m3=56257950/131=429450
  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
    818687750818010
    687758188463101
    81863126201-12
    6362111-1213
    621620-1213-818
    So our multiplicative inverse is 13 mod 818 ≡ 13
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    5251071580525010
    10715852520458101
    525589301-9
    5831911-9172
    3130-9172-525
    So our multiplicative inverse is 172 mod 525 ≡ 172
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    1314294500131010
    429450131327832101
    131324301-4
    3231021-441
    3211-441-45
    212041-45131
    So our multiplicative inverse is -45 mod 131 ≡ 86
    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
    =  (159 × 68775 × 13 +
       175 × 107158 × 172 +
       37 × 429450 × 86)   mod 56257950
    = 8455825 (mod 56257950)


    So our answer is 8455825 (mod 56257950).


Verification

So we found that x ≡ 8455825
If this is correct, then the following statements (i.e. the original equations) are true:
813x (mod 818) ≡ 23 (mod 818)
206x (mod 525) ≡ 875 (mod 525)
263x (mod 131) ≡ 430 (mod 131)

Let's see whether that's indeed the case if we use x ≡ 8455825.