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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
4217256101-5
72611111-56
611156-56-35
116156-3541
6511-3541-76
515041-76421
So our multiplicative inverse is -76 mod 421 ≡ 345
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1315320131010
53213148101
131816301-16
83221-1633
3211-1633-49
212033-49131
So our multiplicative inverse is -49 mod 131 ≡ 82
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1916426301-2
6463111-23
631630-23-191
So our multiplicative inverse is 3 mod 191 ≡ 3
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 ≡ 334 × 72-1 (mod 421) ≡ 334 × 345 (mod 421) ≡ 297 (mod 421)
x ≡ 240 × 532-1 (mod 131) ≡ 240 × 82 (mod 131) ≡ 30 (mod 131)
x ≡ 566 × 64-1 (mod 191) ≡ 566 × 3 (mod 191) ≡ 170 (mod 191)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 421 × 131 × 191 = 10533841
  2. We calculate the numbers M1 to M3
    M1=M/m1=10533841/421=25021,   M2=M/m2=10533841/131=80411,   M3=M/m3=10533841/191=55151
  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
    421250210421010
    2502142159182101
    42118225701-2
    182573111-27
    571152-27-37
    112517-37192
    2120-37192-421
    So our multiplicative inverse is 192 mod 421 ≡ 192
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    131804110131010
    80411131613108101
    13110812301-1
    108234161-15
    231617-15-6
    167225-617
    7231-617-57
    212017-57131
    So our multiplicative inverse is -57 mod 131 ≡ 74
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    191551510191010
    55151191288143101
    19114314801-1
    143482471-13
    484711-13-4
    4714703-4191
    So our multiplicative inverse is -4 mod 191 ≡ 187
    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
    =  (297 × 25021 × 192 +
       30 × 80411 × 74 +
       170 × 55151 × 187)   mod 10533841
    = 8798776 (mod 10533841)


    So our answer is 8798776 (mod 10533841).


Verification

So we found that x ≡ 8798776
If this is correct, then the following statements (i.e. the original equations) are true:
72x (mod 421) ≡ 334 (mod 421)
532x (mod 131) ≡ 240 (mod 131)
64x (mod 191) ≡ 566 (mod 191)

Let's see whether that's indeed the case if we use x ≡ 8798776.