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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
1915940191010
594191321101
191219201-9
2121011-991
2120-991-191
So our multiplicative inverse is 91 mod 191 ≡ 91
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
771547122401-1
5472242991-13
22499226-13-7
99263213-724
262115-724-31
2154124-31148
5150-31148-771
So our multiplicative inverse is 148 mod 771 ≡ 148
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
70120139801-3
20198251-37
985193-37-136
53127-136143
3211-136143-279
2120143-279701
So our multiplicative inverse is -279 mod 701 ≡ 422
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 ≡ 920 × 594-1 (mod 191) ≡ 920 × 91 (mod 191) ≡ 62 (mod 191)
x ≡ 306 × 547-1 (mod 771) ≡ 306 × 148 (mod 771) ≡ 570 (mod 771)
x ≡ 586 × 201-1 (mod 701) ≡ 586 × 422 (mod 701) ≡ 540 (mod 701)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 191 × 771 × 701 = 103229961
  2. We calculate the numbers M1 to M3
    M1=M/m1=103229961/191=540471,   M2=M/m2=103229961/771=133891,   M3=M/m3=103229961/701=147261
  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
    1915404710191010
    5404711912829132101
    19113215901-1
    132592141-13
    591443-13-13
    143423-1355
    3211-1355-68
    212055-68191
    So our multiplicative inverse is -68 mod 191 ≡ 123
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    7711338910771010
    133891771173508101
    771508126301-1
    50826312451-12
    263245118-12-3
    2451813112-341
    181117-341-44
    1171441-4485
    7413-4485-129
    431185-129214
    3130-129214-771
    So our multiplicative inverse is 214 mod 771 ≡ 214
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    7011472610701010
    14726170121051101
    70151133801-13
    51381131-1314
    3813212-1314-41
    13121114-4155
    121120-4155-701
    So our multiplicative inverse is 55 mod 701 ≡ 55
    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
    =  (62 × 540471 × 123 +
       570 × 133891 × 214 +
       540 × 147261 × 55)   mod 103229961
    = 52117086 (mod 103229961)


    So our answer is 52117086 (mod 103229961).


Verification

So we found that x ≡ 52117086
If this is correct, then the following statements (i.e. the original equations) are true:
594x (mod 191) ≡ 920 (mod 191)
547x (mod 771) ≡ 306 (mod 771)
201x (mod 701) ≡ 586 (mod 701)

Let's see whether that's indeed the case if we use x ≡ 52117086.