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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
7187710718010
771718153101
71853132901-13
53291241-1314
292415-1314-27
2454414-27122
5411-27122-149
4140122-149718
So our multiplicative inverse is -149 mod 718 ≡ 569
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
83971083010
971831158101
835812501-1
5825281-13
25831-13-10
81803-1083
So our multiplicative inverse is -10 mod 83 ≡ 73
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1992140199010
214199115101
1991513401-13
154331-1340
4311-1340-53
313040-53199
So our multiplicative inverse is -53 mod 199 ≡ 146
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 ≡ 631 × 771-1 (mod 718) ≡ 631 × 569 (mod 718) ≡ 39 (mod 718)
x ≡ 484 × 971-1 (mod 83) ≡ 484 × 73 (mod 83) ≡ 57 (mod 83)
x ≡ 120 × 214-1 (mod 199) ≡ 120 × 146 (mod 199) ≡ 8 (mod 199)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 718 × 83 × 199 = 11859206
  2. We calculate the numbers M1 to M3
    M1=M/m1=11859206/718=16517,   M2=M/m2=11859206/83=142882,   M3=M/m3=11859206/199=59594
  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
    718165170718010
    16517718233101
    7183239101-239
    31301-239718
    So our multiplicative inverse is -239 mod 718 ≡ 479
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    83142882083010
    14288283172139101
    83392501-2
    395741-215
    5411-215-17
    414015-1783
    So our multiplicative inverse is -17 mod 83 ≡ 66
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    199595940199010
    5959419929993101
    1999321301-2
    9313721-215
    13261-215-92
    212015-92199
    So our multiplicative inverse is -92 mod 199 ≡ 107
    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
    =  (39 × 16517 × 479 +
       57 × 142882 × 66 +
       8 × 59594 × 107)   mod 11859206
    = 7648175 (mod 11859206)


    So our answer is 7648175 (mod 11859206).


Verification

So we found that x ≡ 7648175
If this is correct, then the following statements (i.e. the original equations) are true:
771x (mod 718) ≡ 631 (mod 718)
971x (mod 83) ≡ 484 (mod 83)
214x (mod 199) ≡ 120 (mod 199)

Let's see whether that's indeed the case if we use x ≡ 7648175.