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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
23485023010
48523212101
23211101-11
21201-1123
So our multiplicative inverse is -11 mod 23 ≡ 12
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1247070124010
707124587101
1248713701-1
87372131-13
3713211-13-7
1311123-710
11251-710-57
212010-57124
So our multiplicative inverse is -57 mod 124 ≡ 67
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
7497670749010
767749118101
74918411101-41
1811171-4142
11714-4142-83
741342-83125
4311-83125-208
3130125-208749
So our multiplicative inverse is -208 mod 749 ≡ 541
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 ≡ 888 × 485-1 (mod 23) ≡ 888 × 12 (mod 23) ≡ 7 (mod 23)
x ≡ 623 × 707-1 (mod 124) ≡ 623 × 67 (mod 124) ≡ 77 (mod 124)
x ≡ 110 × 767-1 (mod 749) ≡ 110 × 541 (mod 749) ≡ 339 (mod 749)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 23 × 124 × 749 = 2136148
  2. We calculate the numbers M1 to M3
    M1=M/m1=2136148/23=92876,   M2=M/m2=2136148/124=17227,   M3=M/m3=2136148/749=2852
  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
    2392876023010
    928762340382101
    23211101-11
    21201-1123
    So our multiplicative inverse is -11 mod 23 ≡ 12
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    124172270124010
    17227124138115101
    1241151901-1
    11591271-113
    9712-113-14
    723113-1455
    2120-1455-124
    So our multiplicative inverse is 55 mod 124 ≡ 55
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    74928520749010
    28527493605101
    749605114401-1
    6051444291-15
    14429428-15-21
    2928115-2126
    281280-2126-749
    So our multiplicative inverse is 26 mod 749 ≡ 26
    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
    =  (7 × 92876 × 12 +
       77 × 17227 × 55 +
       339 × 2852 × 26)   mod 2136148
    = 1224205 (mod 2136148)


    So our answer is 1224205 (mod 2136148).


Verification

So we found that x ≡ 1224205
If this is correct, then the following statements (i.e. the original equations) are true:
485x (mod 23) ≡ 888 (mod 23)
707x (mod 124) ≡ 623 (mod 124)
767x (mod 749) ≡ 110 (mod 749)

Let's see whether that's indeed the case if we use x ≡ 1224205.