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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
29546029010
546291824101
29241501-1
245441-15
5411-15-6
41405-629
So our multiplicative inverse is -6 mod 29 ≡ 23
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
82727301701-30
27171101-3031
171017-3031-61
1071331-6192
7321-6192-245
313092-245827
So our multiplicative inverse is -245 mod 827 ≡ 582
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
4678360467010
8364671369101
46736919801-1
369983751-14
9875123-14-5
7523364-519
23635-519-62
651119-6281
5150-6281-467
So our multiplicative inverse is 81 mod 467 ≡ 81
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 ≡ 573 × 546-1 (mod 29) ≡ 573 × 23 (mod 29) ≡ 13 (mod 29)
x ≡ 906 × 27-1 (mod 827) ≡ 906 × 582 (mod 827) ≡ 493 (mod 827)
x ≡ 487 × 836-1 (mod 467) ≡ 487 × 81 (mod 467) ≡ 219 (mod 467)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 29 × 827 × 467 = 11200061
  2. We calculate the numbers M1 to M3
    M1=M/m1=11200061/29=386209,   M2=M/m2=11200061/827=13543,   M3=M/m3=11200061/467=23983
  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
    29386209029010
    386209291331716101
    291611301-1
    1613131-12
    13341-12-9
    31302-929
    So our multiplicative inverse is -9 mod 29 ≡ 20
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    827135430827010
    1354382716311101
    827311220501-2
    31120511061-23
    205106199-23-5
    10699173-58
    997141-58-117
    71708-117827
    So our multiplicative inverse is -117 mod 827 ≡ 710
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    467239830467010
    2398346751166101
    467166213501-2
    1661351311-23
    13531411-23-14
    3111293-1431
    11912-1431-45
    924131-45211
    2120-45211-467
    So our multiplicative inverse is 211 mod 467 ≡ 211
    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 × 386209 × 20 +
       493 × 13543 × 710 +
       219 × 23983 × 211)   mod 11200061
    = 1868686 (mod 11200061)


    So our answer is 1868686 (mod 11200061).


Verification

So we found that x ≡ 1868686
If this is correct, then the following statements (i.e. the original equations) are true:
546x (mod 29) ≡ 573 (mod 29)
27x (mod 827) ≡ 906 (mod 827)
836x (mod 467) ≡ 487 (mod 467)

Let's see whether that's indeed the case if we use x ≡ 1868686.