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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
2039700203010
9702034158101
20315814501-1
158453231-14
4523122-14-5
2322114-59
221220-59-203
So our multiplicative inverse is 9 mod 203 ≡ 9
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
823491133201-1
49133211591-12
332159214-12-5
159141152-557
14524-557-119
541157-119176
4140-119176-823
So our multiplicative inverse is 176 mod 823 ≡ 176
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
73971512401-1
7152429191-130
241915-130-31
1953430-31123
5411-31123-154
4140123-154739
So our multiplicative inverse is -154 mod 739 ≡ 585
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 ≡ 227 × 970-1 (mod 203) ≡ 227 × 9 (mod 203) ≡ 13 (mod 203)
x ≡ 10 × 491-1 (mod 823) ≡ 10 × 176 (mod 823) ≡ 114 (mod 823)
x ≡ 826 × 715-1 (mod 739) ≡ 826 × 585 (mod 739) ≡ 643 (mod 739)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 203 × 823 × 739 = 123463991
  2. We calculate the numbers M1 to M3
    M1=M/m1=123463991/203=608197,   M2=M/m2=123463991/823=150017,   M3=M/m3=123463991/739=167069
  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
    2036081970203010
    60819720329969101
    203922501-22
    95141-2223
    5411-2223-45
    414023-45203
    So our multiplicative inverse is -45 mod 203 ≡ 158
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    8231500170823010
    150017823182231101
    823231313001-3
    23113011011-34
    130101129-34-7
    101293144-725
    291421-725-57
    14114025-57823
    So our multiplicative inverse is -57 mod 823 ≡ 766
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    7391670690739010
    16706973922655101
    73955132401-13
    5524271-1327
    24733-1327-94
    732127-94215
    3130-94215-739
    So our multiplicative inverse is 215 mod 739 ≡ 215
    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
    =  (13 × 608197 × 158 +
       114 × 150017 × 766 +
       643 × 167069 × 215)   mod 123463991
    = 36185778 (mod 123463991)


    So our answer is 36185778 (mod 123463991).


Verification

So we found that x ≡ 36185778
If this is correct, then the following statements (i.e. the original equations) are true:
970x (mod 203) ≡ 227 (mod 203)
491x (mod 823) ≡ 10 (mod 823)
715x (mod 739) ≡ 826 (mod 739)

Let's see whether that's indeed the case if we use x ≡ 36185778.