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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
11332011010
33211302101
1125101-5
21201-511
So our multiplicative inverse is -5 mod 11 ≡ 6
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
3378180337010
8183372144101
33714424901-2
144492461-25
494613-25-7
4631515-7110
3130-7110-337
So our multiplicative inverse is 110 mod 337 ≡ 110
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
4198640419010
864419226101
4192616301-16
263821-16129
3211-16129-145
2120129-145419
So our multiplicative inverse is -145 mod 419 ≡ 274
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 ≡ 114 × 332-1 (mod 11) ≡ 114 × 6 (mod 11) ≡ 2 (mod 11)
x ≡ 443 × 818-1 (mod 337) ≡ 443 × 110 (mod 337) ≡ 202 (mod 337)
x ≡ 543 × 864-1 (mod 419) ≡ 543 × 274 (mod 419) ≡ 37 (mod 419)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 11 × 337 × 419 = 1553233
  2. We calculate the numbers M1 to M3
    M1=M/m1=1553233/11=141203,   M2=M/m2=1553233/337=4609,   M3=M/m3=1553233/419=3707
  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
    11141203011010
    14120311128367101
    1171401-1
    74131-12
    4311-12-3
    31302-311
    So our multiplicative inverse is -3 mod 11 ≡ 8
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    33746090337010
    460933713228101
    337228110901-1
    2281092101-13
    10910109-13-31
    109113-3134
    9190-3134-337
    So our multiplicative inverse is 34 mod 337 ≡ 34
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    41937070419010
    37074198355101
    41935516401-1
    355645351-16
    6435129-16-7
    3529166-713
    29645-713-59
    651113-5972
    5150-5972-419
    So our multiplicative inverse is 72 mod 419 ≡ 72
    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
    =  (2 × 141203 × 8 +
       202 × 4609 × 34 +
       37 × 3707 × 72)   mod 1553233
    = 298784 (mod 1553233)


    So our answer is 298784 (mod 1553233).


Verification

So we found that x ≡ 298784
If this is correct, then the following statements (i.e. the original equations) are true:
332x (mod 11) ≡ 114 (mod 11)
818x (mod 337) ≡ 443 (mod 337)
864x (mod 419) ≡ 543 (mod 419)

Let's see whether that's indeed the case if we use x ≡ 298784.