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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
3674170367010
417367150101
3675071701-7
50172161-715
171611-715-22
16116015-22367
So our multiplicative inverse is -22 mod 367 ≡ 345
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1115770111010
577111522101
111225101-5
2212201-5111
So our multiplicative inverse is -5 mod 111 ≡ 106
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
61779061010
779611247101
614711401-1
4714351-14
14524-14-9
54114-913
4140-913-61
So our multiplicative inverse is 13 mod 61 ≡ 13
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 ≡ 421 × 417-1 (mod 367) ≡ 421 × 345 (mod 367) ≡ 280 (mod 367)
x ≡ 46 × 577-1 (mod 111) ≡ 46 × 106 (mod 111) ≡ 103 (mod 111)
x ≡ 595 × 779-1 (mod 61) ≡ 595 × 13 (mod 61) ≡ 49 (mod 61)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 367 × 111 × 61 = 2484957
  2. We calculate the numbers M1 to M3
    M1=M/m1=2484957/367=6771,   M2=M/m2=2484957/111=22387,   M3=M/m3=2484957/61=40737
  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
    36767710367010
    677136718165101
    36716523701-2
    165374171-29
    371723-29-20
    173529-20109
    3211-20109-129
    2120109-129367
    So our multiplicative inverse is -129 mod 367 ≡ 238
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    111223870111010
    2238711120176101
    1117613501-1
    7635261-13
    35655-13-16
    65113-1619
    5150-1619-111
    So our multiplicative inverse is 19 mod 111 ≡ 19
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    6140737061010
    407376166750101
    615011101-1
    5011461-15
    11615-15-6
    65115-611
    5150-611-61
    So our multiplicative inverse is 11 mod 61 ≡ 11
    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
    =  (280 × 6771 × 238 +
       103 × 22387 × 19 +
       49 × 40737 × 11)   mod 2484957
    = 116986 (mod 2484957)


    So our answer is 116986 (mod 2484957).


Verification

So we found that x ≡ 116986
If this is correct, then the following statements (i.e. the original equations) are true:
417x (mod 367) ≡ 421 (mod 367)
577x (mod 111) ≡ 46 (mod 111)
779x (mod 61) ≡ 595 (mod 61)

Let's see whether that's indeed the case if we use x ≡ 116986.