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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!

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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
1494232301-3
42231191-34
231914-34-7
194434-732
4311-732-39
313032-39149
So our multiplicative inverse is -39 mod 149 ≡ 110
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
631263210501-2
2631052531-25
10553152-25-7
5352115-712
521520-712-631
So our multiplicative inverse is 12 mod 631 ≡ 12
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
637879501-79
85131-7980
5312-7980-159
321180-159239
2120-159239-637
So our multiplicative inverse is 239 mod 637 ≡ 239
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 ≡ 767 × 42-1 (mod 149) ≡ 767 × 110 (mod 149) ≡ 36 (mod 149)
x ≡ 754 × 263-1 (mod 631) ≡ 754 × 12 (mod 631) ≡ 214 (mod 631)
x ≡ 897 × 8-1 (mod 637) ≡ 897 × 239 (mod 637) ≡ 351 (mod 637)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 149 × 631 × 637 = 59890103
  2. We calculate the numbers M1 to M3
    M1=M/m1=59890103/149=401947,   M2=M/m2=59890103/631=94913,   M3=M/m3=59890103/637=94019
  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
    1494019470149010
    401947149269794101
    1499415501-1
    94551391-12
    5539116-12-3
    3916272-38
    16722-38-19
    72318-1965
    2120-1965-149
    So our multiplicative inverse is 65 mod 149 ≡ 65
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    631949130631010
    94913631150263101
    631263210501-2
    2631052531-25
    10553152-25-7
    5352115-712
    521520-712-631
    So our multiplicative inverse is 12 mod 631 ≡ 12
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    637940190637010
    94019637147380101
    637380125701-1
    38025711231-12
    257123211-12-5
    123111122-557
    11251-557-290
    212057-290637
    So our multiplicative inverse is -290 mod 637 ≡ 347
    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
    =  (36 × 401947 × 65 +
       214 × 94913 × 12 +
       351 × 94019 × 347)   mod 59890103
    = 58603077 (mod 59890103)


    So our answer is 58603077 (mod 59890103).


Verification

So we found that x ≡ 58603077
If this is correct, then the following statements (i.e. the original equations) are true:
42x (mod 149) ≡ 767 (mod 149)
263x (mod 631) ≡ 754 (mod 631)
8x (mod 637) ≡ 897 (mod 637)

Let's see whether that's indeed the case if we use x ≡ 58603077.