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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
722207010
2227315101
751201-1
52211-13
2120-13-7
So our multiplicative inverse is 3 mod 7 ≡ 3
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
81541193601-19
4136151-1920
36571-1920-159
515020-159815
So our multiplicative inverse is -159 mod 815 ≡ 656
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
9139180913010
91891315101
9135182301-182
53121-182183
3211-182183-365
2120183-365913
So our multiplicative inverse is -365 mod 913 ≡ 548
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 ≡ 564 × 222-1 (mod 7) ≡ 564 × 3 (mod 7) ≡ 5 (mod 7)
x ≡ 938 × 41-1 (mod 815) ≡ 938 × 656 (mod 815) ≡ 3 (mod 815)
x ≡ 162 × 918-1 (mod 913) ≡ 162 × 548 (mod 913) ≡ 215 (mod 913)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 7 × 815 × 913 = 5208665
  2. We calculate the numbers M1 to M3
    M1=M/m1=5208665/7=744095,   M2=M/m2=5208665/815=6391,   M3=M/m3=5208665/913=5705
  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
    774409507010
    74409571062992101
    723101-3
    21201-37
    So our multiplicative inverse is -3 mod 7 ≡ 4
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    81563910815010
    63918157686101
    815686112901-1
    6861295411-16
    1294136-16-19
    416656-19120
    6511-19120-139
    5150120-139815
    So our multiplicative inverse is -139 mod 815 ≡ 676
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    91357050913010
    57059136227101
    9132274501-4
    22754521-4181
    5221-4181-366
    2120181-366913
    So our multiplicative inverse is -366 mod 913 ≡ 547
    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
    =  (5 × 744095 × 4 +
       3 × 6391 × 676 +
       215 × 5705 × 547)   mod 5208665
    = 818263 (mod 5208665)


    So our answer is 818263 (mod 5208665).


Verification

So we found that x ≡ 818263
If this is correct, then the following statements (i.e. the original equations) are true:
222x (mod 7) ≡ 564 (mod 7)
41x (mod 815) ≡ 938 (mod 815)
918x (mod 913) ≡ 162 (mod 913)

Let's see whether that's indeed the case if we use x ≡ 818263.