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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
523198212701-2
1981271711-23
12771156-23-5
71561153-58
5615311-58-29
1511148-2937
11423-2937-103
431137-103140
3130-103140-523
So our multiplicative inverse is 140 mod 523 ≡ 140
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
3895360389010
5363891147101
38914729501-2
147951521-23
9552143-23-5
5243193-58
43947-58-37
97128-3745
7231-3745-172
212045-172389
So our multiplicative inverse is -172 mod 389 ≡ 217
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
89316059301-5
160931671-56
9367126-56-11
67262156-1128
2615111-1128-39
15111428-3967
11423-3967-173
431167-173240
3130-173240-893
So our multiplicative inverse is 240 mod 893 ≡ 240
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 ≡ 58 × 198-1 (mod 523) ≡ 58 × 140 (mod 523) ≡ 275 (mod 523)
x ≡ 336 × 536-1 (mod 389) ≡ 336 × 217 (mod 389) ≡ 169 (mod 389)
x ≡ 703 × 160-1 (mod 893) ≡ 703 × 240 (mod 893) ≡ 836 (mod 893)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 523 × 389 × 893 = 181678171
  2. We calculate the numbers M1 to M3
    M1=M/m1=181678171/523=347377,   M2=M/m2=181678171/389=467039,   M3=M/m3=181678171/893=203447
  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
    5233473770523010
    347377523664105101
    523105410301-4
    105103121-45
    1032511-45-259
    21205-259523
    So our multiplicative inverse is -259 mod 523 ≡ 264
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    3894670390389010
    4670393891200239101
    389239115001-1
    2391501891-12
    15089161-12-3
    89611282-35
    612825-35-13
    285535-1370
    5312-1370-83
    321170-83153
    2120-83153-389
    So our multiplicative inverse is 153 mod 389 ≡ 153
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    8932034470893010
    203447893227736101
    893736115701-1
    73615741081-15
    157108149-15-6
    108492105-617
    491049-617-74
    1091117-7491
    9190-7491-893
    So our multiplicative inverse is 91 mod 893 ≡ 91
    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
    =  (275 × 347377 × 264 +
       169 × 467039 × 153 +
       836 × 203447 × 91)   mod 181678171
    = 86562005 (mod 181678171)


    So our answer is 86562005 (mod 181678171).


Verification

So we found that x ≡ 86562005
If this is correct, then the following statements (i.e. the original equations) are true:
198x (mod 523) ≡ 58 (mod 523)
536x (mod 389) ≡ 336 (mod 389)
160x (mod 893) ≡ 703 (mod 893)

Let's see whether that's indeed the case if we use x ≡ 86562005.