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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
6629172501-7
91253161-722
251619-722-29
1691722-2951
9712-2951-80
723151-80291
2120-80291-662
So our multiplicative inverse is 291 mod 662 ≡ 291
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
7978510797010
851797154101
79754144101-14
54411131-1415
411332-1415-59
1326115-59369
2120-59369-797
So our multiplicative inverse is 369 mod 797 ≡ 369
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
41934317601-1
343764391-15
7639137-15-6
3937125-611
372181-611-204
212011-204419
So our multiplicative inverse is -204 mod 419 ≡ 215
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 ≡ 657 × 91-1 (mod 662) ≡ 657 × 291 (mod 662) ≡ 531 (mod 662)
x ≡ 599 × 851-1 (mod 797) ≡ 599 × 369 (mod 797) ≡ 262 (mod 797)
x ≡ 130 × 343-1 (mod 419) ≡ 130 × 215 (mod 419) ≡ 296 (mod 419)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 662 × 797 × 419 = 221070266
  2. We calculate the numbers M1 to M3
    M1=M/m1=221070266/662=333943,   M2=M/m2=221070266/797=277378,   M3=M/m3=221070266/419=527614
  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
    6623339430662010
    333943662504295101
    66229527201-2
    29572471-29
    727102-29-92
    72319-92285
    2120-92285-662
    So our multiplicative inverse is 285 mod 662 ≡ 285
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    7972773780797010
    27737879734822101
    7972236501-36
    225421-36145
    5221-36145-326
    2120145-326797
    So our multiplicative inverse is -326 mod 797 ≡ 471
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    4195276140419010
    527614419125993101
    4199344701-4
    93471461-45
    474611-45-9
    4614605-9419
    So our multiplicative inverse is -9 mod 419 ≡ 410
    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
    =  (531 × 333943 × 285 +
       262 × 277378 × 471 +
       296 × 527614 × 410)   mod 221070266
    = 17209883 (mod 221070266)


    So our answer is 17209883 (mod 221070266).


Verification

So we found that x ≡ 17209883
If this is correct, then the following statements (i.e. the original equations) are true:
91x (mod 662) ≡ 657 (mod 662)
851x (mod 797) ≡ 599 (mod 797)
343x (mod 419) ≡ 130 (mod 419)

Let's see whether that's indeed the case if we use x ≡ 17209883.