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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
19815019010
815194217101
19171201-1
172811-19
2120-19-19
So our multiplicative inverse is 9 mod 19 ≡ 9
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
3377160337010
716337242101
337428101-8
4214201-8337
So our multiplicative inverse is -8 mod 337 ≡ 329
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1153260115010
326115296101
1159611901-1
9619511-16
191190-16-115
So our multiplicative inverse is 6 mod 115 ≡ 6
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 ≡ 181 × 815-1 (mod 19) ≡ 181 × 9 (mod 19) ≡ 14 (mod 19)
x ≡ 244 × 716-1 (mod 337) ≡ 244 × 329 (mod 337) ≡ 70 (mod 337)
x ≡ 808 × 326-1 (mod 115) ≡ 808 × 6 (mod 115) ≡ 18 (mod 115)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 19 × 337 × 115 = 736345
  2. We calculate the numbers M1 to M3
    M1=M/m1=736345/19=38755,   M2=M/m2=736345/337=2185,   M3=M/m3=736345/115=6403
  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
    1938755019010
    3875519203914101
    19141501-1
    145241-13
    5411-13-4
    41403-419
    So our multiplicative inverse is -4 mod 19 ≡ 15
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    33721850337010
    21853376163101
    33716321101-2
    163111491-229
    11912-229-31
    924129-31153
    2120-31153-337
    So our multiplicative inverse is 153 mod 337 ≡ 153
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    11564030115010
    64031155578101
    1157813701-1
    7837241-13
    37491-13-28
    41403-28115
    So our multiplicative inverse is -28 mod 115 ≡ 87
    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
    =  (14 × 38755 × 15 +
       70 × 2185 × 153 +
       18 × 6403 × 87)   mod 736345
    = 331678 (mod 736345)


    So our answer is 331678 (mod 736345).


Verification

So we found that x ≡ 331678
If this is correct, then the following statements (i.e. the original equations) are true:
815x (mod 19) ≡ 181 (mod 19)
716x (mod 337) ≡ 244 (mod 337)
326x (mod 115) ≡ 808 (mod 115)

Let's see whether that's indeed the case if we use x ≡ 331678.