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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
4218550421010
855421213101
4211332501-32
135231-3265
5312-3265-97
321165-97162
2120-97162-421
So our multiplicative inverse is 162 mod 421 ≡ 162
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
78911569901-6
115991161-67
991663-67-48
163517-48247
3130-48247-789
So our multiplicative inverse is 247 mod 789 ≡ 247
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
3773037010
7337136101
37361101-1
3613601-137
So our multiplicative inverse is -1 mod 37 ≡ 36
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 ≡ 351 × 855-1 (mod 421) ≡ 351 × 162 (mod 421) ≡ 27 (mod 421)
x ≡ 28 × 115-1 (mod 789) ≡ 28 × 247 (mod 789) ≡ 604 (mod 789)
x ≡ 113 × 73-1 (mod 37) ≡ 113 × 36 (mod 37) ≡ 35 (mod 37)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 421 × 789 × 37 = 12290253
  2. We calculate the numbers M1 to M3
    M1=M/m1=12290253/421=29193,   M2=M/m2=12290253/789=15577,   M3=M/m3=12290253/37=332169
  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
    421291930421010
    2919342169144101
    421144213301-2
    1441331111-23
    13311121-23-38
    1111103-38421
    So our multiplicative inverse is -38 mod 421 ≡ 383
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    789155770789010
    1557778919586101
    789586120301-1
    58620321801-13
    203180123-13-4
    180237193-431
    231914-431-35
    1944331-35171
    4311-35171-206
    3130171-206789
    So our multiplicative inverse is -206 mod 789 ≡ 583
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    37332169037010
    33216937897720101
    372011701-1
    2017131-12
    17352-12-11
    32112-1113
    2120-1113-37
    So our multiplicative inverse is 13 mod 37 ≡ 13
    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
    =  (27 × 29193 × 383 +
       604 × 15577 × 583 +
       35 × 332169 × 13)   mod 12290253
    = 1989673 (mod 12290253)


    So our answer is 1989673 (mod 12290253).


Verification

So we found that x ≡ 1989673
If this is correct, then the following statements (i.e. the original equations) are true:
855x (mod 421) ≡ 351 (mod 421)
115x (mod 789) ≡ 28 (mod 789)
73x (mod 37) ≡ 113 (mod 37)

Let's see whether that's indeed the case if we use x ≡ 1989673.