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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
53396053010
39653725101
53252301-2
253811-217
3130-217-53
So our multiplicative inverse is 17 mod 53 ≡ 17
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1934530193010
453193267101
1936725901-2
6759181-23
59873-23-23
83223-2349
3211-2349-72
212049-72193
So our multiplicative inverse is -72 mod 193 ≡ 121
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
565305010
65351303101
531201-1
32111-12
2120-12-5
So our multiplicative inverse is 2 mod 5 ≡ 2
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 ≡ 916 × 396-1 (mod 53) ≡ 916 × 17 (mod 53) ≡ 43 (mod 53)
x ≡ 721 × 453-1 (mod 193) ≡ 721 × 121 (mod 193) ≡ 5 (mod 193)
x ≡ 891 × 653-1 (mod 5) ≡ 891 × 2 (mod 5) ≡ 2 (mod 5)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 53 × 193 × 5 = 51145
  2. We calculate the numbers M1 to M3
    M1=M/m1=51145/53=965,   M2=M/m2=51145/193=265,   M3=M/m3=51145/5=10229
  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
    53965053010
    965531811101
    53114901-4
    119121-45
    9241-45-24
    21205-2453
    So our multiplicative inverse is -24 mod 53 ≡ 29
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    1932650193010
    265193172101
    1937224901-2
    72491231-23
    492323-23-8
    233723-859
    3211-859-67
    212059-67193
    So our multiplicative inverse is -67 mod 193 ≡ 126
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    51022905010
    10229520454101
    541101-1
    41401-15
    So our multiplicative inverse is -1 mod 5 ≡ 4
    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
    =  (43 × 965 × 29 +
       5 × 265 × 126 +
       2 × 10229 × 4)   mod 51145
    = 20077 (mod 51145)


    So our answer is 20077 (mod 51145).


Verification

So we found that x ≡ 20077
If this is correct, then the following statements (i.e. the original equations) are true:
396x (mod 53) ≡ 916 (mod 53)
453x (mod 193) ≡ 721 (mod 193)
653x (mod 5) ≡ 891 (mod 5)

Let's see whether that's indeed the case if we use x ≡ 20077.