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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
475333114201-1
3331422491-13
14249244-13-7
4944153-710
44584-710-87
541110-8797
4140-8797-475
So our multiplicative inverse is 97 mod 475 ≡ 97
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
13786013010
78613606101
1362101-2
61601-213
So our multiplicative inverse is -2 mod 13 ≡ 11
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
617218218101-2
2181811371-23
18137433-23-14
3733143-1417
33481-1417-150
414017-150617
So our multiplicative inverse is -150 mod 617 ≡ 467
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 ≡ 693 × 333-1 (mod 475) ≡ 693 × 97 (mod 475) ≡ 246 (mod 475)
x ≡ 604 × 786-1 (mod 13) ≡ 604 × 11 (mod 13) ≡ 1 (mod 13)
x ≡ 684 × 218-1 (mod 617) ≡ 684 × 467 (mod 617) ≡ 439 (mod 617)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 475 × 13 × 617 = 3809975
  2. We calculate the numbers M1 to M3
    M1=M/m1=3809975/475=8021,   M2=M/m2=3809975/13=293075,   M3=M/m3=3809975/617=6175
  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
    47580210475010
    802147516421101
    47542115401-1
    421547431-18
    5443111-18-9
    43113108-935
    111011-935-44
    10110035-44475
    So our multiplicative inverse is -44 mod 475 ≡ 431
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    13293075013010
    29307513225443101
    1334101-4
    31301-413
    So our multiplicative inverse is -4 mod 13 ≡ 9
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    61761750617010
    6175617105101
    6175123201-123
    52211-123247
    2120-123247-617
    So our multiplicative inverse is 247 mod 617 ≡ 247
    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
    =  (246 × 8021 × 431 +
       1 × 293075 × 9 +
       439 × 6175 × 247)   mod 3809975
    = 2465971 (mod 3809975)


    So our answer is 2465971 (mod 3809975).


Verification

So we found that x ≡ 2465971
If this is correct, then the following statements (i.e. the original equations) are true:
333x (mod 475) ≡ 693 (mod 475)
786x (mod 13) ≡ 604 (mod 13)
218x (mod 617) ≡ 684 (mod 617)

Let's see whether that's indeed the case if we use x ≡ 2465971.