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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
32312028301-2
120831371-23
833729-23-8
379413-835
9190-835-323
So our multiplicative inverse is 35 mod 323 ≡ 35
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
5871058701-58
107131-5859
7321-5859-176
313059-176587
So our multiplicative inverse is -176 mod 587 ≡ 411
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
895776111901-1
7761196621-17
11962157-17-8
6257157-815
575112-815-173
522115-173361
2120-173361-895
So our multiplicative inverse is 361 mod 895 ≡ 361
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 ≡ 703 × 120-1 (mod 323) ≡ 703 × 35 (mod 323) ≡ 57 (mod 323)
x ≡ 442 × 10-1 (mod 587) ≡ 442 × 411 (mod 587) ≡ 279 (mod 587)
x ≡ 128 × 776-1 (mod 895) ≡ 128 × 361 (mod 895) ≡ 563 (mod 895)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 323 × 587 × 895 = 169692895
  2. We calculate the numbers M1 to M3
    M1=M/m1=169692895/323=525365,   M2=M/m2=169692895/587=289085,   M3=M/m3=169692895/895=189601
  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
    3235253650323010
    5253653231626167101
    323167115601-1
    1671561111-12
    15611142-12-29
    112512-29147
    2120-29147-323
    So our multiplicative inverse is 147 mod 323 ≡ 147
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    5872890850587010
    289085587492281101
    58728122501-2
    281251161-223
    25641-223-94
    616023-94587
    So our multiplicative inverse is -94 mod 587 ≡ 493
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    8951896010895010
    189601895211756101
    895756113901-1
    7561395611-16
    13961217-16-13
    61173106-1345
    171017-1345-58
    1071345-58103
    7321-58103-264
    3130103-264895
    So our multiplicative inverse is -264 mod 895 ≡ 631
    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
    =  (57 × 525365 × 147 +
       279 × 289085 × 493 +
       563 × 189601 × 631)   mod 169692895
    = 32899868 (mod 169692895)


    So our answer is 32899868 (mod 169692895).


Verification

So we found that x ≡ 32899868
If this is correct, then the following statements (i.e. the original equations) are true:
120x (mod 323) ≡ 703 (mod 323)
10x (mod 587) ≡ 442 (mod 587)
776x (mod 895) ≡ 128 (mod 895)

Let's see whether that's indeed the case if we use x ≡ 32899868.