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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
5717530571010
7535711182101
57118232501-3
18225771-322
25734-322-69
741322-6991
4311-6991-160
313091-160571
So our multiplicative inverse is -160 mod 571 ≡ 411
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
719418130101-1
41830111171-12
301117267-12-5
117671502-57
6750117-57-12
50172167-1231
171611-1231-43
16116031-43719
So our multiplicative inverse is -43 mod 719 ≡ 676
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
95290714501-1
907452071-121
45763-121-127
732121-127275
3130-127275-952
So our multiplicative inverse is 275 mod 952 ≡ 275
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 ≡ 766 × 753-1 (mod 571) ≡ 766 × 411 (mod 571) ≡ 205 (mod 571)
x ≡ 76 × 418-1 (mod 719) ≡ 76 × 676 (mod 719) ≡ 327 (mod 719)
x ≡ 464 × 907-1 (mod 952) ≡ 464 × 275 (mod 952) ≡ 32 (mod 952)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 571 × 719 × 952 = 390842648
  2. We calculate the numbers M1 to M3
    M1=M/m1=390842648/571=684488,   M2=M/m2=390842648/719=543592,   M3=M/m3=390842648/952=410549
  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
    5716844880571010
    6844885711198430101
    571430114101-1
    430141371-14
    1417201-14-81
    71704-81571
    So our multiplicative inverse is -81 mod 571 ≡ 490
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    7195435920719010
    54359271975628101
    71928251901-25
    2819191-2526
    19921-2526-77
    919026-77719
    So our multiplicative inverse is -77 mod 719 ≡ 642
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    9524105490952010
    410549952431237101
    9522374401-4
    23745911-4237
    4140-4237-952
    So our multiplicative inverse is 237 mod 952 ≡ 237
    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
    =  (205 × 684488 × 490 +
       327 × 543592 × 642 +
       32 × 410549 × 237)   mod 390842648
    = 338608344 (mod 390842648)


    So our answer is 338608344 (mod 390842648).


Verification

So we found that x ≡ 338608344
If this is correct, then the following statements (i.e. the original equations) are true:
753x (mod 571) ≡ 766 (mod 571)
418x (mod 719) ≡ 76 (mod 719)
907x (mod 952) ≡ 464 (mod 952)

Let's see whether that's indeed the case if we use x ≡ 338608344.