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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
1573310157010
331157217101
157179401-9
174411-937
4140-937-157
So our multiplicative inverse is 37 mod 157 ≡ 37
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
6537300653010
730653177101
6537783701-8
7737231-817
373121-817-212
313017-212653
So our multiplicative inverse is -212 mod 653 ≡ 441
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
43226043010
22643511101
431131001-3
1110111-34
101100-34-43
So our multiplicative inverse is 4 mod 43 ≡ 4
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 ≡ 231 × 331-1 (mod 157) ≡ 231 × 37 (mod 157) ≡ 69 (mod 157)
x ≡ 240 × 730-1 (mod 653) ≡ 240 × 441 (mod 653) ≡ 54 (mod 653)
x ≡ 177 × 226-1 (mod 43) ≡ 177 × 4 (mod 43) ≡ 20 (mod 43)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 157 × 653 × 43 = 4408403
  2. We calculate the numbers M1 to M3
    M1=M/m1=4408403/157=28079,   M2=M/m2=4408403/653=6751,   M3=M/m3=4408403/43=102521
  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
    157280790157010
    28079157178133101
    15713312401-1
    133245131-16
    2413111-16-7
    1311126-713
    11251-713-72
    212013-72157
    So our multiplicative inverse is -72 mod 157 ≡ 85
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    65367510653010
    675165310221101
    653221221101-2
    2212111101-23
    21110211-23-65
    1011003-65653
    So our multiplicative inverse is -65 mod 653 ≡ 588
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    43102521043010
    1025214323849101
    4394701-4
    97121-45
    7231-45-19
    21205-1943
    So our multiplicative inverse is -19 mod 43 ≡ 24
    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
    =  (69 × 28079 × 85 +
       54 × 6751 × 588 +
       20 × 102521 × 24)   mod 4408403
    = 636076 (mod 4408403)


    So our answer is 636076 (mod 4408403).


Verification

So we found that x ≡ 636076
If this is correct, then the following statements (i.e. the original equations) are true:
331x (mod 157) ≡ 231 (mod 157)
730x (mod 653) ≡ 240 (mod 653)
226x (mod 43) ≡ 177 (mod 43)

Let's see whether that's indeed the case if we use x ≡ 636076.