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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
5879250587010
9255871338101
587338124901-1
3382491891-12
24989271-12-5
89711182-57
7118317-57-26
1817117-2633
171170-2633-587
So our multiplicative inverse is 33 mod 587 ≡ 33
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
90129431901-3
294191591-346
19921-346-95
919046-95901
So our multiplicative inverse is -95 mod 901 ≡ 806
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1233640123010
3641232118101
1231181501-1
11852331-124
5312-124-25
321124-2549
2120-2549-123
So our multiplicative inverse is 49 mod 123 ≡ 49
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 ≡ 692 × 925-1 (mod 587) ≡ 692 × 33 (mod 587) ≡ 530 (mod 587)
x ≡ 299 × 294-1 (mod 901) ≡ 299 × 806 (mod 901) ≡ 427 (mod 901)
x ≡ 674 × 364-1 (mod 123) ≡ 674 × 49 (mod 123) ≡ 62 (mod 123)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 587 × 901 × 123 = 65053101
  2. We calculate the numbers M1 to M3
    M1=M/m1=65053101/587=110823,   M2=M/m2=65053101/901=72201,   M3=M/m3=65053101/123=528887
  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
    5871108230587010
    110823587188467101
    587467112001-1
    46712031071-14
    120107113-14-5
    10713834-544
    13341-544-181
    313044-181587
    So our multiplicative inverse is -181 mod 587 ≡ 406
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    901722010901010
    7220190180121101
    90112175401-7
    121542131-715
    541342-715-67
    1326115-67417
    2120-67417-901
    So our multiplicative inverse is 417 mod 901 ≡ 417
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    1235288870123010
    5288871234299110101
    12311011301-1
    11013861-19
    13621-19-19
    61609-19123
    So our multiplicative inverse is -19 mod 123 ≡ 104
    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
    =  (530 × 110823 × 406 +
       427 × 72201 × 417 +
       62 × 528887 × 104)   mod 65053101
    = 40484159 (mod 65053101)


    So our answer is 40484159 (mod 65053101).


Verification

So we found that x ≡ 40484159
If this is correct, then the following statements (i.e. the original equations) are true:
925x (mod 587) ≡ 692 (mod 587)
294x (mod 901) ≡ 299 (mod 901)
364x (mod 123) ≡ 674 (mod 123)

Let's see whether that's indeed the case if we use x ≡ 40484159.