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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
837538129901-1
53829912391-12
299239160-12-3
239603592-311
605911-311-14
59159011-14837
So our multiplicative inverse is -14 mod 837 ≡ 823
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1019880101010
988101979101
1017912201-1
79223131-14
221319-14-5
139144-59
9421-59-23
41409-23101
So our multiplicative inverse is -23 mod 101 ≡ 78
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
49747102701-10
47271201-1011
272017-1011-21
2072611-2153
7611-2153-74
616053-74497
So our multiplicative inverse is -74 mod 497 ≡ 423
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 ≡ 672 × 538-1 (mod 837) ≡ 672 × 823 (mod 837) ≡ 636 (mod 837)
x ≡ 365 × 988-1 (mod 101) ≡ 365 × 78 (mod 101) ≡ 89 (mod 101)
x ≡ 562 × 47-1 (mod 497) ≡ 562 × 423 (mod 497) ≡ 160 (mod 497)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 837 × 101 × 497 = 42014889
  2. We calculate the numbers M1 to M3
    M1=M/m1=42014889/837=50197,   M2=M/m2=42014889/101=415989,   M3=M/m3=42014889/497=84537
  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
    837501970837010
    5019783759814101
    83781412301-1
    814233591-136
    23925-136-73
    951436-73109
    5411-73109-182
    4140109-182837
    So our multiplicative inverse is -182 mod 837 ≡ 655
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    1014159890101010
    415989101411871101
    1017113001-1
    71302111-13
    301128-13-7
    118133-710
    8322-710-27
    321110-2737
    2120-2737-101
    So our multiplicative inverse is 37 mod 101 ≡ 37
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    497845370497010
    8453749717047101
    49747102701-10
    47271201-1011
    272017-1011-21
    2072611-2153
    7611-2153-74
    616053-74497
    So our multiplicative inverse is -74 mod 497 ≡ 423
    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
    =  (636 × 50197 × 655 +
       89 × 415989 × 37 +
       160 × 84537 × 423)   mod 42014889
    = 20466123 (mod 42014889)


    So our answer is 20466123 (mod 42014889).


Verification

So we found that x ≡ 20466123
If this is correct, then the following statements (i.e. the original equations) are true:
538x (mod 837) ≡ 672 (mod 837)
988x (mod 101) ≡ 365 (mod 101)
47x (mod 497) ≡ 562 (mod 497)

Let's see whether that's indeed the case if we use x ≡ 20466123.