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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
7288650728010
8657281137101
72813754301-5
13743381-516
43853-516-85
832216-85186
3211-85186-271
2120186-271728
So our multiplicative inverse is -271 mod 728 ≡ 457
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
2813930281010
3932811112101
28111225701-2
112571551-23
575512-23-5
5522713-5138
2120-5138-281
So our multiplicative inverse is 138 mod 281 ≡ 138
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
781587119401-1
587194351-14
1945384-14-153
54114-153157
4140-153157-781
So our multiplicative inverse is 157 mod 781 ≡ 157
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 ≡ 472 × 865-1 (mod 728) ≡ 472 × 457 (mod 728) ≡ 216 (mod 728)
x ≡ 666 × 393-1 (mod 281) ≡ 666 × 138 (mod 281) ≡ 21 (mod 281)
x ≡ 156 × 587-1 (mod 781) ≡ 156 × 157 (mod 781) ≡ 281 (mod 781)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 728 × 281 × 781 = 159767608
  2. We calculate the numbers M1 to M3
    M1=M/m1=159767608/728=219461,   M2=M/m2=159767608/281=568568,   M3=M/m3=159767608/781=204568
  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
    7282194610728010
    219461728301333101
    72833326201-2
    333625231-211
    6223216-211-24
    23161711-2435
    16722-2435-94
    723135-94317
    2120-94317-728
    So our multiplicative inverse is 317 mod 728 ≡ 317
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    2815685680281010
    5685682812023105101
    28110527101-2
    105711341-23
    713423-23-8
    3431113-891
    3130-891-281
    So our multiplicative inverse is 91 mod 281 ≡ 91
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    7812045680781010
    204568781261727101
    78172715401-1
    7275413251-114
    542524-114-29
    2546114-29188
    4140-29188-781
    So our multiplicative inverse is 188 mod 781 ≡ 188
    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
    =  (216 × 219461 × 317 +
       21 × 568568 × 91 +
       281 × 204568 × 188)   mod 159767608
    = 79427200 (mod 159767608)


    So our answer is 79427200 (mod 159767608).


Verification

So we found that x ≡ 79427200
If this is correct, then the following statements (i.e. the original equations) are true:
865x (mod 728) ≡ 472 (mod 728)
393x (mod 281) ≡ 666 (mod 281)
587x (mod 781) ≡ 156 (mod 781)

Let's see whether that's indeed the case if we use x ≡ 79427200.