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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
43114043010
11443228101
432811501-1
28151131-12
151312-12-3
132612-320
2120-320-43
So our multiplicative inverse is 20 mod 43 ≡ 20
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
737266220501-2
2662051611-23
20561322-23-11
61222173-1125
221715-1125-36
1753225-36133
5221-36133-302
2120133-302737
So our multiplicative inverse is -302 mod 737 ≡ 435
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
3496760349010
6763491327101
34932712201-1
3272214191-115
221913-115-16
1936115-16111
3130-16111-349
So our multiplicative inverse is 111 mod 349 ≡ 111
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 ≡ 877 × 114-1 (mod 43) ≡ 877 × 20 (mod 43) ≡ 39 (mod 43)
x ≡ 705 × 266-1 (mod 737) ≡ 705 × 435 (mod 737) ≡ 83 (mod 737)
x ≡ 306 × 676-1 (mod 349) ≡ 306 × 111 (mod 349) ≡ 113 (mod 349)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 43 × 737 × 349 = 11060159
  2. We calculate the numbers M1 to M3
    M1=M/m1=11060159/43=257213,   M2=M/m2=11060159/737=15007,   M3=M/m3=11060159/349=31691
  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
    43257213043010
    25721343598130101
    433011301-1
    3013241-13
    13431-13-10
    41403-1043
    So our multiplicative inverse is -10 mod 43 ≡ 33
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    737150070737010
    1500773720267101
    737267220301-2
    2672031641-23
    20364311-23-11
    6411593-1158
    11912-1158-69
    924158-69334
    2120-69334-737
    So our multiplicative inverse is 334 mod 737 ≡ 334
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    349316910349010
    3169134990281101
    34928116801-1
    28168491-15
    68975-15-36
    95145-3641
    5411-3641-77
    414041-77349
    So our multiplicative inverse is -77 mod 349 ≡ 272
    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
    =  (39 × 257213 × 33 +
       83 × 15007 × 334 +
       113 × 31691 × 272)   mod 11060159
    = 6787116 (mod 11060159)


    So our answer is 6787116 (mod 11060159).


Verification

So we found that x ≡ 6787116
If this is correct, then the following statements (i.e. the original equations) are true:
114x (mod 43) ≡ 877 (mod 43)
266x (mod 737) ≡ 705 (mod 737)
676x (mod 349) ≡ 306 (mod 349)

Let's see whether that's indeed the case if we use x ≡ 6787116.