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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
1572770157010
2771571120101
15712013701-1
12037391-14
37941-14-17
91904-17157
So our multiplicative inverse is -17 mod 157 ≡ 140
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
37110735001-3
10750271-37
50771-37-52
71707-52371
So our multiplicative inverse is -52 mod 371 ≡ 319
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
5575630557010
56355716101
557692501-92
65111-9293
5150-9293-557
So our multiplicative inverse is 93 mod 557 ≡ 93
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 ≡ 774 × 277-1 (mod 157) ≡ 774 × 140 (mod 157) ≡ 30 (mod 157)
x ≡ 941 × 107-1 (mod 371) ≡ 941 × 319 (mod 371) ≡ 40 (mod 371)
x ≡ 339 × 563-1 (mod 557) ≡ 339 × 93 (mod 557) ≡ 335 (mod 557)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 157 × 371 × 557 = 32443579
  2. We calculate the numbers M1 to M3
    M1=M/m1=32443579/157=206647,   M2=M/m2=32443579/371=87449,   M3=M/m3=32443579/557=58247
  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
    1572066470157010
    206647157131635101
    1573541701-4
    3517211-49
    171170-49-157
    So our multiplicative inverse is 9 mod 157 ≡ 9
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    371874490371010
    87449371235264101
    371264110701-1
    2641072501-13
    1075027-13-7
    507713-752
    7170-752-371
    So our multiplicative inverse is 52 mod 371 ≡ 52
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    557582470557010
    58247557104319101
    557319123801-1
    3192381811-12
    23881276-12-5
    8176152-57
    765151-57-110
    51507-110557
    So our multiplicative inverse is -110 mod 557 ≡ 447
    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 × 206647 × 9 +
       40 × 87449 × 52 +
       335 × 58247 × 447)   mod 32443579
    = 5457821 (mod 32443579)


    So our answer is 5457821 (mod 32443579).


Verification

So we found that x ≡ 5457821
If this is correct, then the following statements (i.e. the original equations) are true:
277x (mod 157) ≡ 774 (mod 157)
107x (mod 371) ≡ 941 (mod 371)
563x (mod 557) ≡ 339 (mod 557)

Let's see whether that's indeed the case if we use x ≡ 5457821.