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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
2299040229010
9042293217101
22921711201-1
217121811-119
121120-119-229
So our multiplicative inverse is 19 mod 229 ≡ 19
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1935870193010
58719338101
193824101-24
81801-24193
So our multiplicative inverse is -24 mod 193 ≡ 169
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
2518540251010
8542513101101
25110124901-2
10149231-25
493161-25-82
31305-82251
So our multiplicative inverse is -82 mod 251 ≡ 169
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 ≡ 952 × 904-1 (mod 229) ≡ 952 × 19 (mod 229) ≡ 226 (mod 229)
x ≡ 594 × 587-1 (mod 193) ≡ 594 × 169 (mod 193) ≡ 26 (mod 193)
x ≡ 146 × 854-1 (mod 251) ≡ 146 × 169 (mod 251) ≡ 76 (mod 251)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 229 × 193 × 251 = 11093447
  2. We calculate the numbers M1 to M3
    M1=M/m1=11093447/229=48443,   M2=M/m2=11093447/193=57479,   M3=M/m3=11093447/251=44197
  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
    229484430229010
    48443229211124101
    229124110501-1
    1241051191-12
    10519510-12-11
    1910192-1113
    10911-1113-24
    919013-24229
    So our multiplicative inverse is -24 mod 229 ≡ 205
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    193574790193010
    57479193297158101
    19315813501-1
    158354181-15
    3518117-15-6
    1817115-611
    171170-611-193
    So our multiplicative inverse is 11 mod 193 ≡ 11
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    251441970251010
    4419725117621101
    25121112001-11
    2120111-1112
    201200-1112-251
    So our multiplicative inverse is 12 mod 251 ≡ 12
    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
    =  (226 × 48443 × 205 +
       26 × 57479 × 11 +
       76 × 44197 × 12)   mod 11093447
    = 4767319 (mod 11093447)


    So our answer is 4767319 (mod 11093447).


Verification

So we found that x ≡ 4767319
If this is correct, then the following statements (i.e. the original equations) are true:
904x (mod 229) ≡ 952 (mod 229)
587x (mod 193) ≡ 594 (mod 193)
854x (mod 251) ≡ 146 (mod 251)

Let's see whether that's indeed the case if we use x ≡ 4767319.