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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
5794114501-14
415811-14113
5150-14113-579
So our multiplicative inverse is 113 mod 579 ≡ 113
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
541394114701-1
39414721001-13
147100147-13-4
10047263-411
47675-411-81
651111-8192
5150-8192-541
So our multiplicative inverse is 92 mod 541 ≡ 92
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
53392053010
39253721101
532121101-2
21111101-23
111011-23-5
1011003-553
So our multiplicative inverse is -5 mod 53 ≡ 48
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 ≡ 974 × 41-1 (mod 579) ≡ 974 × 113 (mod 579) ≡ 52 (mod 579)
x ≡ 122 × 394-1 (mod 541) ≡ 122 × 92 (mod 541) ≡ 404 (mod 541)
x ≡ 125 × 392-1 (mod 53) ≡ 125 × 48 (mod 53) ≡ 11 (mod 53)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 579 × 541 × 53 = 16601667
  2. We calculate the numbers M1 to M3
    M1=M/m1=16601667/579=28673,   M2=M/m2=16601667/541=30687,   M3=M/m3=16601667/53=313239
  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
    579286730579010
    2867357949302101
    579302127701-1
    3022771251-12
    27725112-12-23
    2521212-23278
    2120-23278-579
    So our multiplicative inverse is 278 mod 579 ≡ 278
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    541306870541010
    3068754156391101
    541391115001-1
    3911502911-13
    15091159-13-4
    91591323-47
    5932127-47-11
    3227157-1118
    27552-1118-101
    522118-101220
    2120-101220-541
    So our multiplicative inverse is 220 mod 541 ≡ 220
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    53313239053010
    3132395359109101
    5395801-5
    98111-56
    8180-56-53
    So our multiplicative inverse is 6 mod 53 ≡ 6
    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
    =  (52 × 28673 × 278 +
       404 × 30687 × 220 +
       11 × 313239 × 6)   mod 16601667
    = 8314492 (mod 16601667)


    So our answer is 8314492 (mod 16601667).


Verification

So we found that x ≡ 8314492
If this is correct, then the following statements (i.e. the original equations) are true:
41x (mod 579) ≡ 974 (mod 579)
394x (mod 541) ≡ 122 (mod 541)
392x (mod 53) ≡ 125 (mod 53)

Let's see whether that's indeed the case if we use x ≡ 8314492.