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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
7038960703010
8967031193101
703193312401-3
1931241691-34
12469155-34-7
69551144-711
5514313-711-40
14131111-4051
131130-4051-703
So our multiplicative inverse is 51 mod 703 ≡ 51
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
3219550321010
9553212313101
3213131801-1
31383911-140
8180-140-321
So our multiplicative inverse is 40 mod 321 ≡ 40
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
43521043010
52143125101
4358301-8
53121-89
3211-89-17
21209-1743
So our multiplicative inverse is -17 mod 43 ≡ 26
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 ≡ 582 × 896-1 (mod 703) ≡ 582 × 51 (mod 703) ≡ 156 (mod 703)
x ≡ 797 × 955-1 (mod 321) ≡ 797 × 40 (mod 321) ≡ 101 (mod 321)
x ≡ 571 × 521-1 (mod 43) ≡ 571 × 26 (mod 43) ≡ 11 (mod 43)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 703 × 321 × 43 = 9703509
  2. We calculate the numbers M1 to M3
    M1=M/m1=9703509/703=13803,   M2=M/m2=9703509/321=30229,   M3=M/m3=9703509/43=225663
  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
    703138030703010
    1380370319446101
    703446125701-1
    44625711891-12
    257189168-12-3
    189682532-38
    6853115-38-11
    5315388-1141
    15817-1141-52
    871141-5293
    7170-5293-703
    So our multiplicative inverse is 93 mod 703 ≡ 93
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    321302290321010
    302293219455101
    3215554601-5
    5546191-56
    46951-56-35
    91906-35321
    So our multiplicative inverse is -35 mod 321 ≡ 286
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    43225663043010
    22566343524742101
    43421101-1
    4214201-143
    So our multiplicative inverse is -1 mod 43 ≡ 42
    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
    =  (156 × 13803 × 93 +
       101 × 30229 × 286 +
       11 × 225663 × 42)   mod 9703509
    = 3580535 (mod 9703509)


    So our answer is 3580535 (mod 9703509).


Verification

So we found that x ≡ 3580535
If this is correct, then the following statements (i.e. the original equations) are true:
896x (mod 703) ≡ 582 (mod 703)
955x (mod 321) ≡ 797 (mod 321)
521x (mod 43) ≡ 571 (mod 43)

Let's see whether that's indeed the case if we use x ≡ 3580535.