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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
53157053010
15753251101
53511201-1
5122511-126
2120-126-53
So our multiplicative inverse is 26 mod 53 ≡ 26
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
54953811101-1
5381148101-149
111011-149-50
10110049-50549
So our multiplicative inverse is -50 mod 549 ≡ 499
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
44310542301-4
105234131-417
2313110-417-21
13101317-2138
10331-2138-135
313038-135443
So our multiplicative inverse is -135 mod 443 ≡ 308
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 ≡ 499 × 157-1 (mod 53) ≡ 499 × 26 (mod 53) ≡ 42 (mod 53)
x ≡ 45 × 538-1 (mod 549) ≡ 45 × 499 (mod 549) ≡ 495 (mod 549)
x ≡ 661 × 105-1 (mod 443) ≡ 661 × 308 (mod 443) ≡ 251 (mod 443)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 53 × 549 × 443 = 12889971
  2. We calculate the numbers M1 to M3
    M1=M/m1=12889971/53=243207,   M2=M/m2=12889971/549=23479,   M3=M/m3=12889971/443=29097
  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
    53243207053010
    24320753458843101
    534311001-1
    4310431-15
    10331-15-16
    31305-1653
    So our multiplicative inverse is -16 mod 53 ≡ 37
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    549234790549010
    2347954942421101
    549421112801-1
    4211283371-14
    12837317-14-13
    3717234-1330
    17352-1330-163
    321130-163193
    2120-163193-549
    So our multiplicative inverse is 193 mod 549 ≡ 193
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    443290970443010
    2909744365302101
    443302114101-1
    3021412201-13
    1412071-13-22
    2012003-22443
    So our multiplicative inverse is -22 mod 443 ≡ 421
    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
    =  (42 × 243207 × 37 +
       495 × 23479 × 193 +
       251 × 29097 × 421)   mod 12889971
    = 11241819 (mod 12889971)


    So our answer is 11241819 (mod 12889971).


Verification

So we found that x ≡ 11241819
If this is correct, then the following statements (i.e. the original equations) are true:
157x (mod 53) ≡ 499 (mod 53)
538x (mod 549) ≡ 45 (mod 549)
105x (mod 443) ≡ 661 (mod 443)

Let's see whether that's indeed the case if we use x ≡ 11241819.