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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
83915954401-5
159443271-516
4427117-516-21
271711016-2137
171017-2137-58
1071337-5895
7321-5895-248
313095-248839
So our multiplicative inverse is -248 mod 839 ≡ 591
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
65418065010
41865628101
65282901-2
289311-27
9190-27-65
So our multiplicative inverse is 7 mod 65 ≡ 7
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
4537820453010
7824531329101
453329112401-1
3291242811-13
12481143-13-4
81431383-47
433815-47-11
385737-1184
5312-1184-95
321184-95179
2120-95179-453
So our multiplicative inverse is 179 mod 453 ≡ 179
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 ≡ 350 × 159-1 (mod 839) ≡ 350 × 591 (mod 839) ≡ 456 (mod 839)
x ≡ 12 × 418-1 (mod 65) ≡ 12 × 7 (mod 65) ≡ 19 (mod 65)
x ≡ 147 × 782-1 (mod 453) ≡ 147 × 179 (mod 453) ≡ 39 (mod 453)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 839 × 65 × 453 = 24704355
  2. We calculate the numbers M1 to M3
    M1=M/m1=24704355/839=29445,   M2=M/m2=24704355/65=380067,   M3=M/m3=24704355/453=54535
  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
    839294450839010
    294458393580101
    83980103901-10
    8039221-1021
    392191-1021-409
    212021-409839
    So our multiplicative inverse is -409 mod 839 ≡ 430
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    65380067065010
    38006765584712101
    65125501-5
    125221-511
    5221-511-27
    212011-2765
    So our multiplicative inverse is -27 mod 65 ≡ 38
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    453545350453010
    54535453120175101
    453175210301-2
    1751031721-23
    10372131-23-5
    72312103-513
    311031-513-44
    10110013-44453
    So our multiplicative inverse is -44 mod 453 ≡ 409
    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
    =  (456 × 29445 × 430 +
       19 × 380067 × 38 +
       39 × 54535 × 409)   mod 24704355
    = 652359 (mod 24704355)


    So our answer is 652359 (mod 24704355).


Verification

So we found that x ≡ 652359
If this is correct, then the following statements (i.e. the original equations) are true:
159x (mod 839) ≡ 350 (mod 839)
418x (mod 65) ≡ 12 (mod 65)
782x (mod 453) ≡ 147 (mod 453)

Let's see whether that's indeed the case if we use x ≡ 652359.