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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
39137012101-1
3702117131-118
211318-118-19
1381518-1937
8513-1937-56
531237-5693
3211-5693-149
212093-149391
So our multiplicative inverse is -149 mod 391 ≡ 242
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
6076394001-9
63401231-910
4023117-910-19
23171610-1929
17625-1929-77
651129-77106
5150-77106-607
So our multiplicative inverse is 106 mod 607 ≡ 106
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1633820163010
382163256101
1635625101-2
5651151-23
515101-23-32
51503-32163
So our multiplicative inverse is -32 mod 163 ≡ 131
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 ≡ 201 × 370-1 (mod 391) ≡ 201 × 242 (mod 391) ≡ 158 (mod 391)
x ≡ 256 × 63-1 (mod 607) ≡ 256 × 106 (mod 607) ≡ 428 (mod 607)
x ≡ 454 × 382-1 (mod 163) ≡ 454 × 131 (mod 163) ≡ 142 (mod 163)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 391 × 607 × 163 = 38685931
  2. We calculate the numbers M1 to M3
    M1=M/m1=38685931/391=98941,   M2=M/m2=38685931/607=63733,   M3=M/m3=38685931/163=237337
  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
    391989410391010
    9894139125318101
    39118211301-21
    1813151-2122
    13523-2122-65
    531222-6587
    3211-6587-152
    212087-152391
    So our multiplicative inverse is -152 mod 391 ≡ 239
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    607637330607010
    63733607104605101
    6076051201-1
    605230211-1303
    2120-1303-607
    So our multiplicative inverse is 303 mod 607 ≡ 303
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    1632373370163010
    23733716314569101
    163918101-18
    91901-18163
    So our multiplicative inverse is -18 mod 163 ≡ 145
    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
    =  (158 × 98941 × 239 +
       428 × 63733 × 303 +
       142 × 237337 × 145)   mod 38685931
    = 21063328 (mod 38685931)


    So our answer is 21063328 (mod 38685931).


Verification

So we found that x ≡ 21063328
If this is correct, then the following statements (i.e. the original equations) are true:
370x (mod 391) ≡ 201 (mod 391)
63x (mod 607) ≡ 256 (mod 607)
382x (mod 163) ≡ 454 (mod 163)

Let's see whether that's indeed the case if we use x ≡ 21063328.