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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
1037190103010
7191036101101
1031011201-1
10125011-151
2120-151-103
So our multiplicative inverse is 51 mod 103 ≡ 51
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
4756170475010
6174751142101
47514234901-3
142492441-37
494415-37-10
445847-1087
5411-1087-97
414087-97475
So our multiplicative inverse is -97 mod 475 ≡ 378
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
6617140661010
714661153101
66153122501-12
5325231-1225
25381-1225-212
313025-212661
So our multiplicative inverse is -212 mod 661 ≡ 449
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 ≡ 766 × 719-1 (mod 103) ≡ 766 × 51 (mod 103) ≡ 29 (mod 103)
x ≡ 652 × 617-1 (mod 475) ≡ 652 × 378 (mod 475) ≡ 406 (mod 475)
x ≡ 402 × 714-1 (mod 661) ≡ 402 × 449 (mod 661) ≡ 45 (mod 661)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 103 × 475 × 661 = 32339425
  2. We calculate the numbers M1 to M3
    M1=M/m1=32339425/103=313975,   M2=M/m2=32339425/475=68083,   M3=M/m3=32339425/661=48925
  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
    1033139750103010
    313975103304831101
    1033131001-3
    3110311-310
    101100-310-103
    So our multiplicative inverse is 10 mod 103 ≡ 10
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    475680830475010
    68083475143158101
    4751583101-3
    158115801-3475
    So our multiplicative inverse is -3 mod 475 ≡ 472
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    661489250661010
    489256617411101
    6611160101-60
    1111101-60661
    So our multiplicative inverse is -60 mod 661 ≡ 601
    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
    =  (29 × 313975 × 10 +
       406 × 68083 × 472 +
       45 × 48925 × 601)   mod 32339425
    = 5387856 (mod 32339425)


    So our answer is 5387856 (mod 32339425).


Verification

So we found that x ≡ 5387856
If this is correct, then the following statements (i.e. the original equations) are true:
719x (mod 103) ≡ 766 (mod 103)
617x (mod 475) ≡ 652 (mod 475)
714x (mod 661) ≡ 402 (mod 661)

Let's see whether that's indeed the case if we use x ≡ 5387856.