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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
55249055010
24955429101
552912601-1
2926131-12
26382-12-17
32112-1719
2120-1719-55
So our multiplicative inverse is 19 mod 55 ≡ 19
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
5338220533010
8225331289101
533289124401-1
2892441451-12
24445519-12-11
4519272-1124
19725-1124-59
751224-5983
5221-5983-225
212083-225533
So our multiplicative inverse is -225 mod 533 ≡ 308
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
787501128601-1
50128612151-12
286215171-12-3
21571322-311
712351-311-388
212011-388787
So our multiplicative inverse is -388 mod 787 ≡ 399
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 ≡ 42 × 249-1 (mod 55) ≡ 42 × 19 (mod 55) ≡ 28 (mod 55)
x ≡ 164 × 822-1 (mod 533) ≡ 164 × 308 (mod 533) ≡ 410 (mod 533)
x ≡ 974 × 501-1 (mod 787) ≡ 974 × 399 (mod 787) ≡ 635 (mod 787)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 55 × 533 × 787 = 23070905
  2. We calculate the numbers M1 to M3
    M1=M/m1=23070905/55=419471,   M2=M/m2=23070905/533=43285,   M3=M/m3=23070905/787=29315
  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
    55419471055010
    41947155762641101
    554111401-1
    41142131-13
    141311-13-4
    1311303-455
    So our multiplicative inverse is -4 mod 55 ≡ 51
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    533432850533010
    4328553381112101
    53311248501-4
    112851271-45
    852734-45-19
    274635-19119
    4311-19119-138
    3130119-138533
    So our multiplicative inverse is -138 mod 533 ≡ 395
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    787293150787010
    2931578737196101
    7871964301-4
    19636511-4261
    3130-4261-787
    So our multiplicative inverse is 261 mod 787 ≡ 261
    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
    =  (28 × 419471 × 51 +
       410 × 43285 × 395 +
       635 × 29315 × 261)   mod 23070905
    = 9243163 (mod 23070905)


    So our answer is 9243163 (mod 23070905).


Verification

So we found that x ≡ 9243163
If this is correct, then the following statements (i.e. the original equations) are true:
249x (mod 55) ≡ 42 (mod 55)
822x (mod 533) ≡ 164 (mod 533)
501x (mod 787) ≡ 974 (mod 787)

Let's see whether that's indeed the case if we use x ≡ 9243163.