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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
58349219101-1
492915371-16
9137217-16-13
3717236-1332
17352-1332-173
321132-173205
2120-173205-583
So our multiplicative inverse is 205 mod 583 ≡ 205
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
311173113801-1
1731381351-12
13835333-12-7
3533122-79
332161-79-151
21209-151311
So our multiplicative inverse is -151 mod 311 ≡ 160
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
81838081010
838811028101
812822501-2
2825131-23
25381-23-26
31303-2681
So our multiplicative inverse is -26 mod 81 ≡ 55
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 ≡ 175 × 492-1 (mod 583) ≡ 175 × 205 (mod 583) ≡ 312 (mod 583)
x ≡ 361 × 173-1 (mod 311) ≡ 361 × 160 (mod 311) ≡ 225 (mod 311)
x ≡ 578 × 838-1 (mod 81) ≡ 578 × 55 (mod 81) ≡ 38 (mod 81)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 583 × 311 × 81 = 14686353
  2. We calculate the numbers M1 to M3
    M1=M/m1=14686353/583=25191,   M2=M/m2=14686353/311=47223,   M3=M/m3=14686353/81=181313
  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
    583251910583010
    2519158343122101
    58312249501-4
    122951271-45
    9527314-45-19
    27141135-1924
    141311-1924-43
    13113024-43583
    So our multiplicative inverse is -43 mod 583 ≡ 540
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    311472230311010
    47223311151262101
    31126214901-1
    262495171-16
    4917215-16-13
    1715126-1319
    15271-1319-146
    212019-146311
    So our multiplicative inverse is -146 mod 311 ≡ 165
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    81181313081010
    18131381223835101
    813521101-2
    3511321-27
    11251-27-37
    21207-3781
    So our multiplicative inverse is -37 mod 81 ≡ 44
    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
    =  (312 × 25191 × 540 +
       225 × 47223 × 165 +
       38 × 181313 × 44)   mod 14686353
    = 43454 (mod 14686353)


    So our answer is 43454 (mod 14686353).


Verification

So we found that x ≡ 43454
If this is correct, then the following statements (i.e. the original equations) are true:
492x (mod 583) ≡ 175 (mod 583)
173x (mod 311) ≡ 361 (mod 311)
838x (mod 81) ≡ 578 (mod 81)

Let's see whether that's indeed the case if we use x ≡ 43454.