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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
4516640451010
6644511213101
45121322501-2
213258131-217
2513112-217-19
13121117-1936
121120-1936-451
So our multiplicative inverse is 36 mod 451 ≡ 36
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
77334827701-2
348774401-29
7740137-29-11
4037139-1120
373121-1120-251
313020-251773
So our multiplicative inverse is -251 mod 773 ≡ 522
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1573990157010
399157285101
1578517201-1
85721131-12
721357-12-11
137162-1113
7611-1113-24
616013-24157
So our multiplicative inverse is -24 mod 157 ≡ 133
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 ≡ 670 × 664-1 (mod 451) ≡ 670 × 36 (mod 451) ≡ 217 (mod 451)
x ≡ 172 × 348-1 (mod 773) ≡ 172 × 522 (mod 773) ≡ 116 (mod 773)
x ≡ 662 × 399-1 (mod 157) ≡ 662 × 133 (mod 157) ≡ 126 (mod 157)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 451 × 773 × 157 = 54733811
  2. We calculate the numbers M1 to M3
    M1=M/m1=54733811/451=121361,   M2=M/m2=54733811/773=70807,   M3=M/m3=54733811/157=348623
  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
    4511213610451010
    12136145126942101
    45142103101-10
    42311111-1011
    311129-1011-32
    1191211-3243
    9241-3243-204
    212043-204451
    So our multiplicative inverse is -204 mod 451 ≡ 247
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    773708070773010
    7080777391464101
    773464130901-1
    46430911551-12
    3091551154-12-3
    155154112-35
    15411540-35-773
    So our multiplicative inverse is 5 mod 773 ≡ 5
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    1573486230157010
    348623157222083101
    1578317401-1
    8374191-12
    74982-12-17
    92412-1770
    2120-1770-157
    So our multiplicative inverse is 70 mod 157 ≡ 70
    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
    =  (217 × 121361 × 247 +
       116 × 70807 × 5 +
       126 × 348623 × 70)   mod 54733811
    = 42334234 (mod 54733811)


    So our answer is 42334234 (mod 54733811).


Verification

So we found that x ≡ 42334234
If this is correct, then the following statements (i.e. the original equations) are true:
664x (mod 451) ≡ 670 (mod 451)
348x (mod 773) ≡ 172 (mod 773)
399x (mod 157) ≡ 662 (mod 157)

Let's see whether that's indeed the case if we use x ≡ 42334234.