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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
8212713801-3
27183371-3100
8711-3100-103
7170100-103821
So our multiplicative inverse is -103 mod 821 ≡ 718
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
131401010
31413140101
So our multiplicative inverse is 0 mod 1 ≡ 0
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
7878650787010
865787178101
7877810701-10
7871111-10111
7170-10111-787
So our multiplicative inverse is 111 mod 787 ≡ 111
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 ≡ 99 × 271-1 (mod 821) ≡ 99 × 718 (mod 821) ≡ 476 (mod 821)
x ≡ 284 × 314-1 (mod 1) ≡ 284 × 0 (mod 1) ≡ 0 (mod 1)
x ≡ 438 × 865-1 (mod 787) ≡ 438 × 111 (mod 787) ≡ 611 (mod 787)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 821 × 1 × 787 = 646127
  2. We calculate the numbers M1 to M3
    M1=M/m1=646127/821=787,   M2=M/m2=646127/1=646127,   M3=M/m3=646127/787=821
  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
    82178713401-1
    787342351-124
    34564-124-145
    541124-145169
    4140-145169-821
    So our multiplicative inverse is 169 mod 821 ≡ 169
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    164612701010
    64612716461270101
    So our multiplicative inverse is 0 mod 1 ≡ 0
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    7878210787010
    821787134101
    7873423501-23
    345641-23139
    5411-23139-162
    4140139-162787
    So our multiplicative inverse is -162 mod 787 ≡ 625
    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
    =  (476 × 787 × 169 +
       0 × 646127 × 0 +
       611 × 821 × 625)   mod 646127
    = 136762 (mod 646127)


    So our answer is 136762 (mod 646127).


Verification

So we found that x ≡ 136762
If this is correct, then the following statements (i.e. the original equations) are true:
271x (mod 821) ≡ 99 (mod 821)
314x (mod 1) ≡ 284 (mod 1)
865x (mod 787) ≡ 438 (mod 787)

Let's see whether that's indeed the case if we use x ≡ 136762.