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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
3238700323010
8703232224101
32322419901-1
224992261-13
9926321-13-10
2621153-1013
21541-1013-62
515013-62323
So our multiplicative inverse is -62 mod 323 ≡ 261
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1038320103010
83210388101
103812701-12
87111-1213
7170-1213-103
So our multiplicative inverse is 13 mod 103 ≡ 13
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
2715900271010
590271248101
2714853101-5
48311171-56
3117114-56-11
1714136-1117
14342-1117-79
321117-7996
2120-7996-271
So our multiplicative inverse is 96 mod 271 ≡ 96
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 ≡ 28 × 870-1 (mod 323) ≡ 28 × 261 (mod 323) ≡ 202 (mod 323)
x ≡ 198 × 832-1 (mod 103) ≡ 198 × 13 (mod 103) ≡ 102 (mod 103)
x ≡ 626 × 590-1 (mod 271) ≡ 626 × 96 (mod 271) ≡ 205 (mod 271)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 323 × 103 × 271 = 9015899
  2. We calculate the numbers M1 to M3
    M1=M/m1=9015899/323=27913,   M2=M/m2=9015899/103=87533,   M3=M/m3=9015899/271=33269
  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
    323279130323010
    2791332386135101
    32313525301-2
    135532291-25
    5329124-25-7
    2924155-712
    24544-712-55
    541112-5567
    4140-5567-323
    So our multiplicative inverse is 67 mod 323 ≡ 67
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    103875330103010
    8753310384986101
    1038611701-1
    8617511-16
    171170-16-103
    So our multiplicative inverse is 6 mod 103 ≡ 6
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    271332690271010
    33269271122207101
    27120716401-1
    207643151-14
    641544-14-17
    154334-1755
    4311-1755-72
    313055-72271
    So our multiplicative inverse is -72 mod 271 ≡ 199
    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
    =  (202 × 27913 × 67 +
       102 × 87533 × 6 +
       205 × 33269 × 199)   mod 9015899
    = 3405591 (mod 9015899)


    So our answer is 3405591 (mod 9015899).


Verification

So we found that x ≡ 3405591
If this is correct, then the following statements (i.e. the original equations) are true:
870x (mod 323) ≡ 28 (mod 323)
832x (mod 103) ≡ 198 (mod 103)
590x (mod 271) ≡ 626 (mod 271)

Let's see whether that's indeed the case if we use x ≡ 3405591.