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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
8178892501-9
88253131-928
2513112-928-37
13121128-3765
121120-3765-817
So our multiplicative inverse is 65 mod 817 ≡ 65
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
73932429101-2
324913511-27
9151140-27-9
51401117-916
401137-916-57
1171416-5773
7413-5773-130
431173-130203
3130-130203-739
So our multiplicative inverse is 203 mod 739 ≡ 203
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
59350718601-1
507865771-16
867719-16-7
779856-762
9514-762-69
541162-69131
4140-69131-593
So our multiplicative inverse is 131 mod 593 ≡ 131
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 ≡ 288 × 88-1 (mod 817) ≡ 288 × 65 (mod 817) ≡ 746 (mod 817)
x ≡ 706 × 324-1 (mod 739) ≡ 706 × 203 (mod 739) ≡ 691 (mod 739)
x ≡ 114 × 507-1 (mod 593) ≡ 114 × 131 (mod 593) ≡ 109 (mod 593)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 817 × 739 × 593 = 358031459
  2. We calculate the numbers M1 to M3
    M1=M/m1=358031459/817=438227,   M2=M/m2=358031459/739=484481,   M3=M/m3=358031459/593=603763
  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
    8174382270817010
    438227817536315101
    817315218701-2
    31518711281-23
    187128159-23-5
    128592103-513
    591059-513-70
    1091113-7083
    9190-7083-817
    So our multiplicative inverse is 83 mod 817 ≡ 83
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    7394844810739010
    484481739655436101
    739436130301-1
    43630311331-12
    303133237-12-5
    133373222-517
    3722115-517-22
    22151717-2239
    15721-2239-100
    717039-100739
    So our multiplicative inverse is -100 mod 739 ≡ 639
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    5936037630593010
    603763593101889101
    5938965901-6
    89591301-67
    5930129-67-13
    3029117-1320
    291290-1320-593
    So our multiplicative inverse is 20 mod 593 ≡ 20
    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
    =  (746 × 438227 × 83 +
       691 × 484481 × 639 +
       109 × 603763 × 20)   mod 358031459
    = 343177511 (mod 358031459)


    So our answer is 343177511 (mod 358031459).


Verification

So we found that x ≡ 343177511
If this is correct, then the following statements (i.e. the original equations) are true:
88x (mod 817) ≡ 288 (mod 817)
324x (mod 739) ≡ 706 (mod 739)
507x (mod 593) ≡ 114 (mod 593)

Let's see whether that's indeed the case if we use x ≡ 343177511.