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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
74364419901-1
644996501-17
9950149-17-8
5049117-815
491490-815-743
So our multiplicative inverse is 15 mod 743 ≡ 15
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
871535133601-1
53533611991-12
3361991137-12-3
1991371622-35
13762213-35-13
62134105-1357
131013-1357-70
1033157-70267
3130-70267-871
So our multiplicative inverse is 267 mod 871 ≡ 267
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1494910149010
491149344101
1494431701-3
44172101-37
171017-37-10
107137-1017
7321-1017-44
313017-44149
So our multiplicative inverse is -44 mod 149 ≡ 105
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 ≡ 85 × 644-1 (mod 743) ≡ 85 × 15 (mod 743) ≡ 532 (mod 743)
x ≡ 247 × 535-1 (mod 871) ≡ 247 × 267 (mod 871) ≡ 624 (mod 871)
x ≡ 249 × 491-1 (mod 149) ≡ 249 × 105 (mod 149) ≡ 70 (mod 149)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 743 × 871 × 149 = 96425797
  2. We calculate the numbers M1 to M3
    M1=M/m1=96425797/743=129779,   M2=M/m2=96425797/871=110707,   M3=M/m3=96425797/149=647153
  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
    7431297790743010
    129779743174497101
    743497124601-1
    497246251-13
    2465491-13-148
    51503-148743
    So our multiplicative inverse is -148 mod 743 ≡ 595
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    8711107070871010
    11070787112790101
    8719096101-9
    90611291-910
    612923-910-29
    2939210-29271
    3211-29271-300
    2120271-300871
    So our multiplicative inverse is -300 mod 871 ≡ 571
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    1496471530149010
    647153149434346101
    1494631101-3
    4611421-313
    11251-313-68
    212013-68149
    So our multiplicative inverse is -68 mod 149 ≡ 81
    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
    =  (532 × 129779 × 595 +
       624 × 110707 × 571 +
       70 × 647153 × 81)   mod 96425797
    = 15228317 (mod 96425797)


    So our answer is 15228317 (mod 96425797).


Verification

So we found that x ≡ 15228317
If this is correct, then the following statements (i.e. the original equations) are true:
644x (mod 743) ≡ 85 (mod 743)
535x (mod 871) ≡ 247 (mod 871)
491x (mod 149) ≡ 249 (mod 149)

Let's see whether that's indeed the case if we use x ≡ 15228317.