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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
79428079010
42879533101
793321301-2
3313271-25
13716-25-7
76115-712
6160-712-79
So our multiplicative inverse is 12 mod 79 ≡ 12
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
48941917001-1
419705691-16
706911-16-7
6916906-7489
So our multiplicative inverse is -7 mod 489 ≡ 482
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
2634750263010
4752631212101
26321215101-1
21251481-15
51863-15-31
83225-3167
3211-3167-98
212067-98263
So our multiplicative inverse is -98 mod 263 ≡ 165
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 ≡ 824 × 428-1 (mod 79) ≡ 824 × 12 (mod 79) ≡ 13 (mod 79)
x ≡ 31 × 419-1 (mod 489) ≡ 31 × 482 (mod 489) ≡ 272 (mod 489)
x ≡ 619 × 475-1 (mod 263) ≡ 619 × 165 (mod 263) ≡ 91 (mod 263)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 79 × 489 × 263 = 10159953
  2. We calculate the numbers M1 to M3
    M1=M/m1=10159953/79=128607,   M2=M/m2=10159953/489=20777,   M3=M/m3=10159953/263=38631
  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
    79128607079010
    12860779162774101
    79741501-1
    7451441-115
    5411-115-16
    414015-1679
    So our multiplicative inverse is -16 mod 79 ≡ 63
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    489207770489010
    2077748942239101
    48923921101-2
    239112181-243
    11813-243-45
    832243-45133
    3211-45133-178
    2120133-178489
    So our multiplicative inverse is -178 mod 489 ≡ 311
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    263386310263010
    38631263146233101
    26323313001-1
    233307231-18
    302317-18-9
    237328-935
    7231-935-114
    212035-114263
    So our multiplicative inverse is -114 mod 263 ≡ 149
    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
    =  (13 × 128607 × 63 +
       272 × 20777 × 311 +
       91 × 38631 × 149)   mod 10159953
    = 9265844 (mod 10159953)


    So our answer is 9265844 (mod 10159953).


Verification

So we found that x ≡ 9265844
If this is correct, then the following statements (i.e. the original equations) are true:
428x (mod 79) ≡ 824 (mod 79)
419x (mod 489) ≡ 31 (mod 489)
475x (mod 263) ≡ 619 (mod 263)

Let's see whether that's indeed the case if we use x ≡ 9265844.