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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
841549129201-1
54929212571-12
292257135-12-3
257357122-323
3512211-323-49
12111123-4972
111110-4972-841
So our multiplicative inverse is 72 mod 841 ≡ 72
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
5797730579010
7735791194101
579194219101-2
194191131-23
1913632-23-191
32113-191194
2120-191194-579
So our multiplicative inverse is 194 mod 579 ≡ 194
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
4678990467010
8994671432101
46743213501-1
4323512121-113
3512211-113-27
12111113-2740
111110-2740-467
So our multiplicative inverse is 40 mod 467 ≡ 40
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 ≡ 534 × 549-1 (mod 841) ≡ 534 × 72 (mod 841) ≡ 603 (mod 841)
x ≡ 854 × 773-1 (mod 579) ≡ 854 × 194 (mod 579) ≡ 82 (mod 579)
x ≡ 167 × 899-1 (mod 467) ≡ 167 × 40 (mod 467) ≡ 142 (mod 467)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 841 × 579 × 467 = 227400513
  2. We calculate the numbers M1 to M3
    M1=M/m1=227400513/841=270393,   M2=M/m2=227400513/579=392747,   M3=M/m3=227400513/467=486939
  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
    8412703930841010
    270393841321432101
    841432140901-1
    4324091231-12
    409231718-12-35
    2318152-3537
    18533-3537-146
    531237-146183
    3211-146183-329
    2120183-329841
    So our multiplicative inverse is -329 mod 841 ≡ 512
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    5793927470579010
    392747579678185101
    57918532401-3
    185247171-322
    241717-322-25
    1772322-2572
    7321-2572-169
    313072-169579
    So our multiplicative inverse is -169 mod 579 ≡ 410
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    4674869390467010
    4869394671042325101
    467325114201-1
    3251422411-13
    14241319-13-10
    4119233-1023
    19361-1023-148
    313023-148467
    So our multiplicative inverse is -148 mod 467 ≡ 319
    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
    =  (603 × 270393 × 512 +
       82 × 392747 × 410 +
       142 × 486939 × 319)   mod 227400513
    = 38502424 (mod 227400513)


    So our answer is 38502424 (mod 227400513).


Verification

So we found that x ≡ 38502424
If this is correct, then the following statements (i.e. the original equations) are true:
549x (mod 841) ≡ 534 (mod 841)
773x (mod 579) ≡ 854 (mod 579)
899x (mod 467) ≡ 167 (mod 467)

Let's see whether that's indeed the case if we use x ≡ 38502424.