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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
1676200167010
6201673119101
16711914801-1
119482231-13
482322-13-7
2321113-780
2120-780-167
So our multiplicative inverse is 80 mod 167 ≡ 80
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
53845718101-1
457815521-16
8152129-16-7
52291236-713
292316-713-20
2363513-2073
6511-2073-93
515073-93538
So our multiplicative inverse is -93 mod 538 ≡ 445
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
2015440201010
5442012142101
20114215901-1
142592241-13
5924211-13-7
2411223-717
11251-717-92
212017-92201
So our multiplicative inverse is -92 mod 201 ≡ 109
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 ≡ 258 × 620-1 (mod 167) ≡ 258 × 80 (mod 167) ≡ 99 (mod 167)
x ≡ 2 × 457-1 (mod 538) ≡ 2 × 445 (mod 538) ≡ 352 (mod 538)
x ≡ 13 × 544-1 (mod 201) ≡ 13 × 109 (mod 201) ≡ 10 (mod 201)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 167 × 538 × 201 = 18059046
  2. We calculate the numbers M1 to M3
    M1=M/m1=18059046/167=108138,   M2=M/m2=18059046/538=33567,   M3=M/m3=18059046/201=89846
  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
    1671081380167010
    10813816764789101
    1678917801-1
    89781111-12
    781171-12-15
    1111102-15167
    So our multiplicative inverse is -15 mod 167 ≡ 152
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    538335670538010
    3356753862211101
    538211211601-2
    2111161951-23
    11695121-23-5
    95214113-523
    2111110-523-28
    11101123-2851
    101100-2851-538
    So our multiplicative inverse is 51 mod 538 ≡ 51
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    201898460201010
    89846201446200101
    2012001101-1
    200120001-1201
    So our multiplicative inverse is -1 mod 201 ≡ 200
    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
    =  (99 × 108138 × 152 +
       352 × 33567 × 51 +
       10 × 89846 × 200)   mod 18059046
    = 7694290 (mod 18059046)


    So our answer is 7694290 (mod 18059046).


Verification

So we found that x ≡ 7694290
If this is correct, then the following statements (i.e. the original equations) are true:
620x (mod 167) ≡ 258 (mod 167)
457x (mod 538) ≡ 2 (mod 538)
544x (mod 201) ≡ 13 (mod 201)

Let's see whether that's indeed the case if we use x ≡ 7694290.