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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
97742097010
74297763101
976313401-1
63341291-12
342915-12-3
295542-317
5411-317-20
414017-2097
So our multiplicative inverse is -20 mod 97 ≡ 77
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
2263490226010
3492261123101
226123110301-1
1231031201-12
1032053-12-11
203622-1168
3211-1168-79
212068-79226
So our multiplicative inverse is -79 mod 226 ≡ 147
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
19476019010
47619251101
19119001-19
So our multiplicative inverse is 1 mod 19 ≡ 1
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 ≡ 57 × 742-1 (mod 97) ≡ 57 × 77 (mod 97) ≡ 24 (mod 97)
x ≡ 184 × 349-1 (mod 226) ≡ 184 × 147 (mod 226) ≡ 154 (mod 226)
x ≡ 2 × 476-1 (mod 19) ≡ 2 × 1 (mod 19) ≡ 2 (mod 19)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 97 × 226 × 19 = 416518
  2. We calculate the numbers M1 to M3
    M1=M/m1=416518/97=4294,   M2=M/m2=416518/226=1843,   M3=M/m3=416518/19=21922
  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
    974294097010
    4294974426101
    972631901-3
    2619171-34
    19725-34-11
    75124-1115
    5221-1115-41
    212015-4197
    So our multiplicative inverse is -41 mod 97 ≡ 56
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    22618430226010
    1843226835101
    2263561601-6
    3516231-613
    16351-613-71
    313013-71226
    So our multiplicative inverse is -71 mod 226 ≡ 155
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    1921922019010
    2192219115315101
    19151401-1
    154331-14
    4311-14-5
    31304-519
    So our multiplicative inverse is -5 mod 19 ≡ 14
    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
    =  (24 × 4294 × 56 +
       154 × 1843 × 155 +
       2 × 21922 × 14)   mod 416518
    = 395202 (mod 416518)


    So our answer is 395202 (mod 416518).


Verification

So we found that x ≡ 395202
If this is correct, then the following statements (i.e. the original equations) are true:
742x (mod 97) ≡ 57 (mod 97)
349x (mod 226) ≡ 184 (mod 226)
476x (mod 19) ≡ 2 (mod 19)

Let's see whether that's indeed the case if we use x ≡ 395202.