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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!

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Are you ready to view a full step-by-step Chinese Remainder Theorem calculation for the numbers you have entered? Then use this button!

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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
3018830301010
8833012281101
30128112001-1
281201411-115
201200-115-301
So our multiplicative inverse is 15 mod 301 ≡ 15
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
5298090529010
8095291280101
529280124901-1
2802491311-12
2493181-12-17
3113102-17529
So our multiplicative inverse is -17 mod 529 ≡ 512
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
73463073010
46373625101
732522301-2
2523121-23
232111-23-35
21203-3573
So our multiplicative inverse is -35 mod 73 ≡ 38
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 ≡ 904 × 883-1 (mod 301) ≡ 904 × 15 (mod 301) ≡ 15 (mod 301)
x ≡ 270 × 809-1 (mod 529) ≡ 270 × 512 (mod 529) ≡ 171 (mod 529)
x ≡ 512 × 463-1 (mod 73) ≡ 512 × 38 (mod 73) ≡ 38 (mod 73)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 301 × 529 × 73 = 11623717
  2. We calculate the numbers M1 to M3
    M1=M/m1=11623717/301=38617,   M2=M/m2=11623717/529=21973,   M3=M/m3=11623717/73=159229
  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
    301386170301010
    3861730112889101
    3018933401-3
    89342211-37
    3421113-37-10
    2113187-1017
    13815-1017-27
    851317-2744
    5312-2744-71
    321144-71115
    2120-71115-301
    So our multiplicative inverse is 115 mod 301 ≡ 115
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    529219730529010
    2197352941284101
    529284124501-1
    2842451391-12
    24539611-12-13
    3911362-1341
    11615-1341-54
    651141-5495
    5150-5495-529
    So our multiplicative inverse is 95 mod 529 ≡ 95
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    73159229073010
    15922973218116101
    73164901-4
    169171-45
    9712-45-9
    72315-932
    2120-932-73
    So our multiplicative inverse is 32 mod 73 ≡ 32
    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
    =  (15 × 38617 × 115 +
       171 × 21973 × 95 +
       38 × 159229 × 32)   mod 11623717
    = 1131173 (mod 11623717)


    So our answer is 1131173 (mod 11623717).


Verification

So we found that x ≡ 1131173
If this is correct, then the following statements (i.e. the original equations) are true:
883x (mod 301) ≡ 904 (mod 301)
809x (mod 529) ≡ 270 (mod 529)
463x (mod 73) ≡ 512 (mod 73)

Let's see whether that's indeed the case if we use x ≡ 1131173.