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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
8429985001-8
99501491-89
504911-89-17
4914909-17842
So our multiplicative inverse is -17 mod 842 ≡ 825
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1973552201-5
35221131-56
221319-56-11
139146-1117
9421-1117-45
414017-45197
So our multiplicative inverse is -45 mod 197 ≡ 152
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
2357930235010
793235388101
2358825901-2
88591291-23
592921-23-8
2912903-8235
So our multiplicative inverse is -8 mod 235 ≡ 227
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 ≡ 528 × 99-1 (mod 842) ≡ 528 × 825 (mod 842) ≡ 286 (mod 842)
x ≡ 898 × 35-1 (mod 197) ≡ 898 × 152 (mod 197) ≡ 172 (mod 197)
x ≡ 234 × 793-1 (mod 235) ≡ 234 × 227 (mod 235) ≡ 8 (mod 235)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 842 × 197 × 235 = 38980390
  2. We calculate the numbers M1 to M3
    M1=M/m1=38980390/842=46295,   M2=M/m2=38980390/197=197870,   M3=M/m3=38980390/235=165874
  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
    842462950842010
    4629584254827101
    84282711501-1
    827155521-156
    15271-156-393
    212056-393842
    So our multiplicative inverse is -393 mod 842 ≡ 449
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    1971978700197010
    197870197100482101
    1978223301-2
    82332161-25
    331621-25-12
    1611605-12197
    So our multiplicative inverse is -12 mod 197 ≡ 185
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    2351658740235010
    165874235705199101
    23519913601-1
    199365191-16
    3619117-16-7
    1917126-713
    17281-713-111
    212013-111235
    So our multiplicative inverse is -111 mod 235 ≡ 124
    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
    =  (286 × 46295 × 449 +
       172 × 197870 × 185 +
       8 × 165874 × 124)   mod 38980390
    = 9932518 (mod 38980390)


    So our answer is 9932518 (mod 38980390).


Verification

So we found that x ≡ 9932518
If this is correct, then the following statements (i.e. the original equations) are true:
99x (mod 842) ≡ 528 (mod 842)
35x (mod 197) ≡ 898 (mod 197)
793x (mod 235) ≡ 234 (mod 235)

Let's see whether that's indeed the case if we use x ≡ 9932518.