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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!

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Are you ready to view a full step-by-step Chinese Remainder Theorem calculation for the numbers you have entered? Then use this button!

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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
4479430447010
943447249101
447499601-9
496811-973
6160-973-447
So our multiplicative inverse is 73 mod 447 ≡ 73
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1678590167010
859167524101
1672462301-6
2423111-67
231230-67-167
So our multiplicative inverse is 7 mod 167 ≡ 7
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
91142426301-2
424636461-213
6346117-213-15
461721213-1543
171215-1543-58
1252243-58159
5221-58159-376
2120159-376911
So our multiplicative inverse is -376 mod 911 ≡ 535
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 ≡ 504 × 943-1 (mod 447) ≡ 504 × 73 (mod 447) ≡ 138 (mod 447)
x ≡ 725 × 859-1 (mod 167) ≡ 725 × 7 (mod 167) ≡ 65 (mod 167)
x ≡ 953 × 424-1 (mod 911) ≡ 953 × 535 (mod 911) ≡ 606 (mod 911)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 447 × 167 × 911 = 68005239
  2. We calculate the numbers M1 to M3
    M1=M/m1=68005239/447=152137,   M2=M/m2=68005239/167=407217,   M3=M/m3=68005239/911=74649
  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
    4471521370447010
    152137447340157101
    447157213301-2
    1571331241-23
    13324513-23-17
    24131113-1720
    131112-1720-37
    1125120-37205
    2120-37205-447
    So our multiplicative inverse is 205 mod 447 ≡ 205
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    1674072170167010
    407217167243871101
    1677122501-2
    71252211-25
    252114-25-7
    214515-740
    4140-740-167
    So our multiplicative inverse is 40 mod 167 ≡ 40
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    911746490911010
    7464991181858101
    91185815301-1
    8585316101-117
    531053-117-86
    1033117-86275
    3130-86275-911
    So our multiplicative inverse is 275 mod 911 ≡ 275
    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
    =  (138 × 152137 × 205 +
       65 × 407217 × 40 +
       606 × 74649 × 275)   mod 68005239
    = 53608401 (mod 68005239)


    So our answer is 53608401 (mod 68005239).


Verification

So we found that x ≡ 53608401
If this is correct, then the following statements (i.e. the original equations) are true:
943x (mod 447) ≡ 504 (mod 447)
859x (mod 167) ≡ 725 (mod 167)
424x (mod 911) ≡ 953 (mod 911)

Let's see whether that's indeed the case if we use x ≡ 53608401.