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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
97316097010
31697325101
972532201-3
2522131-34
22371-34-31
31304-3197
So our multiplicative inverse is -31 mod 97 ≡ 66
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
2278980227010
8982273217101
22721711001-1
217102171-122
10713-122-23
732122-2368
3130-2368-227
So our multiplicative inverse is 68 mod 227 ≡ 68
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
148801010
48814880101
So our multiplicative inverse is 0 mod 1 ≡ 0
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 ≡ 265 × 316-1 (mod 97) ≡ 265 × 66 (mod 97) ≡ 30 (mod 97)
x ≡ 709 × 898-1 (mod 227) ≡ 709 × 68 (mod 227) ≡ 88 (mod 227)
x ≡ 490 × 488-1 (mod 1) ≡ 490 × 0 (mod 1) ≡ 0 (mod 1)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 97 × 227 × 1 = 22019
  2. We calculate the numbers M1 to M3
    M1=M/m1=22019/97=227,   M2=M/m2=22019/227=97,   M3=M/m3=22019/1=22019
  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
    97227097010
    22797233101
    973323101-2
    3331121-23
    312151-23-47
    21203-4797
    So our multiplicative inverse is -47 mod 97 ≡ 50
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    2279723301-2
    97332311-25
    333112-25-7
    3121515-7110
    2120-7110-227
    So our multiplicative inverse is 110 mod 227 ≡ 110
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    12201901010
    220191220190101
    So our multiplicative inverse is 0 mod 1 ≡ 0
    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
    =  (30 × 227 × 50 +
       88 × 97 × 110 +
       0 × 22019 × 0)   mod 22019
    = 2358 (mod 22019)


    So our answer is 2358 (mod 22019).


Verification

So we found that x ≡ 2358
If this is correct, then the following statements (i.e. the original equations) are true:
316x (mod 97) ≡ 265 (mod 97)
898x (mod 227) ≡ 709 (mod 227)
488x (mod 1) ≡ 490 (mod 1)

Let's see whether that's indeed the case if we use x ≡ 2358.