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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
21815316501-1
153652231-13
6523219-13-7
2319143-710
19443-710-47
431110-4757
3130-4757-218
So our multiplicative inverse is 57 mod 218 ≡ 57
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
44511839101-3
118911271-34
9127310-34-15
2710274-1534
10713-1534-49
732134-49132
3130-49132-445
So our multiplicative inverse is 132 mod 445 ≡ 132
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
89920647501-4
206752561-49
7556119-49-13
56192189-1335
191811-1335-48
18118035-48899
So our multiplicative inverse is -48 mod 899 ≡ 851
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 ≡ 648 × 153-1 (mod 218) ≡ 648 × 57 (mod 218) ≡ 94 (mod 218)
x ≡ 606 × 118-1 (mod 445) ≡ 606 × 132 (mod 445) ≡ 337 (mod 445)
x ≡ 257 × 206-1 (mod 899) ≡ 257 × 851 (mod 899) ≡ 250 (mod 899)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 218 × 445 × 899 = 87211990
  2. We calculate the numbers M1 to M3
    M1=M/m1=87211990/218=400055,   M2=M/m2=87211990/445=195982,   M3=M/m3=87211990/899=97010
  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
    2184000550218010
    400055218183525101
    2182581801-8
    2518171-89
    18724-89-26
    74139-2635
    4311-2635-61
    313035-61218
    So our multiplicative inverse is -61 mod 218 ≡ 157
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    4451959820445010
    195982445440182101
    44518228101-2
    182812201-25
    812041-25-22
    2012005-22445
    So our multiplicative inverse is -22 mod 445 ≡ 423
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    899970100899010
    97010899107817101
    89981718201-1
    817829791-110
    827913-110-11
    79326110-11296
    3130-11296-899
    So our multiplicative inverse is 296 mod 899 ≡ 296
    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
    =  (94 × 400055 × 157 +
       337 × 195982 × 423 +
       250 × 97010 × 296)   mod 87211990
    = 30546472 (mod 87211990)


    So our answer is 30546472 (mod 87211990).


Verification

So we found that x ≡ 30546472
If this is correct, then the following statements (i.e. the original equations) are true:
153x (mod 218) ≡ 648 (mod 218)
118x (mod 445) ≡ 606 (mod 445)
206x (mod 899) ≡ 257 (mod 899)

Let's see whether that's indeed the case if we use x ≡ 30546472.