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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
74016946401-4
169642411-49
6441123-49-13
41231189-1322
231815-1322-35
1853322-35127
5312-35127-162
3211127-162289
2120-162289-740
So our multiplicative inverse is 289 mod 740 ≡ 289
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
4238560423010
856423210101
4231042301-42
103311-42127
3130-42127-423
So our multiplicative inverse is 127 mod 423 ≡ 127
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1376670137010
6671374119101
13711911801-1
119186111-17
181117-17-8
117147-815
7413-815-23
431115-2338
3130-2338-137
So our multiplicative inverse is 38 mod 137 ≡ 38
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 ≡ 925 × 169-1 (mod 740) ≡ 925 × 289 (mod 740) ≡ 185 (mod 740)
x ≡ 589 × 856-1 (mod 423) ≡ 589 × 127 (mod 423) ≡ 355 (mod 423)
x ≡ 559 × 667-1 (mod 137) ≡ 559 × 38 (mod 137) ≡ 7 (mod 137)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 740 × 423 × 137 = 42883740
  2. We calculate the numbers M1 to M3
    M1=M/m1=42883740/740=57951,   M2=M/m2=42883740/423=101380,   M3=M/m3=42883740/137=313020
  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
    740579510740010
    5795174078231101
    74023134701-3
    231474431-313
    474314-313-16
    43410313-16173
    4311-16173-189
    3130173-189740
    So our multiplicative inverse is -189 mod 740 ≡ 551
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    4231013800423010
    101380423239283101
    423283114001-1
    283140231-13
    1403462-13-139
    32113-139142
    2120-139142-423
    So our multiplicative inverse is 142 mod 423 ≡ 142
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    1373130200137010
    3130201372284112101
    13711212501-1
    112254121-15
    251221-15-11
    1211205-11137
    So our multiplicative inverse is -11 mod 137 ≡ 126
    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
    =  (185 × 57951 × 551 +
       355 × 101380 × 142 +
       7 × 313020 × 126)   mod 42883740
    = 15461005 (mod 42883740)


    So our answer is 15461005 (mod 42883740).


Verification

So we found that x ≡ 15461005
If this is correct, then the following statements (i.e. the original equations) are true:
169x (mod 740) ≡ 925 (mod 740)
856x (mod 423) ≡ 589 (mod 423)
667x (mod 137) ≡ 559 (mod 137)

Let's see whether that's indeed the case if we use x ≡ 15461005.