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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!

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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
961729123201-1
7292323331-14
2323371-14-29
3313304-29961
So our multiplicative inverse is -29 mod 961 ≡ 932
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
35515225101-2
152512501-25
515011-25-7
5015005-7355
So our multiplicative inverse is -7 mod 355 ≡ 348
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
81118746301-4
187632611-49
636112-49-13
6123019-13399
2120-13399-811
So our multiplicative inverse is 399 mod 811 ≡ 399
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 ≡ 289 × 729-1 (mod 961) ≡ 289 × 932 (mod 961) ≡ 268 (mod 961)
x ≡ 522 × 152-1 (mod 355) ≡ 522 × 348 (mod 355) ≡ 251 (mod 355)
x ≡ 641 × 187-1 (mod 811) ≡ 641 × 399 (mod 811) ≡ 294 (mod 811)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 961 × 355 × 811 = 276676705
  2. We calculate the numbers M1 to M3
    M1=M/m1=276676705/961=287905,   M2=M/m2=276676705/355=779371,   M3=M/m3=276676705/811=341155
  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
    9612879050961010
    287905961299566101
    961566139501-1
    56639511711-12
    395171253-12-5
    171533122-517
    531245-517-73
    1252217-73163
    5221-73163-399
    2120163-399961
    So our multiplicative inverse is -399 mod 961 ≡ 562
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    3557793710355010
    7793713552195146101
    35514626301-2
    146632201-25
    632033-25-17
    203625-17107
    3211-17107-124
    2120107-124355
    So our multiplicative inverse is -124 mod 355 ≡ 231
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    8113411550811010
    341155811420535101
    811535127601-1
    53527612591-12
    276259117-12-3
    259171542-347
    17441-347-191
    414047-191811
    So our multiplicative inverse is -191 mod 811 ≡ 620
    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
    =  (268 × 287905 × 562 +
       251 × 779371 × 231 +
       294 × 341155 × 620)   mod 276676705
    = 225415311 (mod 276676705)


    So our answer is 225415311 (mod 276676705).


Verification

So we found that x ≡ 225415311
If this is correct, then the following statements (i.e. the original equations) are true:
729x (mod 961) ≡ 289 (mod 961)
152x (mod 355) ≡ 522 (mod 355)
187x (mod 811) ≡ 641 (mod 811)

Let's see whether that's indeed the case if we use x ≡ 225415311.