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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
22317215101-1
172513191-14
5119213-14-9
1913164-913
13621-913-35
616013-35223
So our multiplicative inverse is -35 mod 223 ≡ 188
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
33409033010
409331213101
33132701-2
137161-23
7611-23-5
61603-533
So our multiplicative inverse is -5 mod 33 ≡ 28
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
5235330523010
533523110101
5231052301-52
103311-52157
3130-52157-523
So our multiplicative inverse is 157 mod 523 ≡ 157
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 ≡ 364 × 172-1 (mod 223) ≡ 364 × 188 (mod 223) ≡ 194 (mod 223)
x ≡ 652 × 409-1 (mod 33) ≡ 652 × 28 (mod 33) ≡ 7 (mod 33)
x ≡ 638 × 533-1 (mod 523) ≡ 638 × 157 (mod 523) ≡ 273 (mod 523)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 223 × 33 × 523 = 3848757
  2. We calculate the numbers M1 to M3
    M1=M/m1=3848757/223=17259,   M2=M/m2=3848757/33=116629,   M3=M/m3=3848757/523=7359
  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
    223172590223010
    172592237788101
    2238824701-2
    88471411-23
    474116-23-5
    416653-533
    6511-533-38
    515033-38223
    So our multiplicative inverse is -38 mod 223 ≡ 185
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    33116629033010
    1166293335347101
    3374501-4
    75121-45
    5221-45-14
    21205-1433
    So our multiplicative inverse is -14 mod 33 ≡ 19
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    52373590523010
    73595231437101
    5233714501-14
    375721-1499
    5221-1499-212
    212099-212523
    So our multiplicative inverse is -212 mod 523 ≡ 311
    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
    =  (194 × 17259 × 185 +
       7 × 116629 × 19 +
       273 × 7359 × 311)   mod 3848757
    = 1194805 (mod 3848757)


    So our answer is 1194805 (mod 3848757).


Verification

So we found that x ≡ 1194805
If this is correct, then the following statements (i.e. the original equations) are true:
172x (mod 223) ≡ 364 (mod 223)
409x (mod 33) ≡ 652 (mod 33)
533x (mod 523) ≡ 638 (mod 523)

Let's see whether that's indeed the case if we use x ≡ 1194805.